# 9.3 Double-angle, half-angle, and reduction formulas  (Page 5/8)

 Page 5 / 8

If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=-8,$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant IV.

For the following exercises, find the values of the six trigonometric functions if the conditions provided hold.

$\mathrm{cos}\left(2\theta \right)=\frac{3}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}90°\le \theta \le 180°$

$\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{2\sqrt{5}}{5},\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{\sqrt{5}}{5},\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{1}{2},\mathrm{csc}\text{\hspace{0.17em}}\theta =\sqrt{5},\mathrm{sec}\text{\hspace{0.17em}}\theta =-\frac{\sqrt{5}}{2},\mathrm{cot}\text{\hspace{0.17em}}\theta =-2$

$\mathrm{cos}\left(2\theta \right)=\frac{1}{\sqrt{2}}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}180°\le \theta \le 270°$

For the following exercises, simplify to one trigonometric expression.

$2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{4}\right)\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\pi }{4}\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{2}\right)$

$4\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{8}\right)\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\pi }{8}\right)$

For the following exercises, find the exact value using half-angle formulas.

$\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{8}\right)\text{\hspace{0.17em}}$

$\frac{\sqrt{2-\sqrt{2}}}{2}$

$\mathrm{cos}\left(-\frac{11\pi }{12}\right)$

$\mathrm{sin}\left(\frac{11\pi }{12}\right)$

$\frac{\sqrt{2-\sqrt{3}}}{2}$

$\mathrm{cos}\left(\frac{7\pi }{8}\right)$

$\mathrm{tan}\left(\frac{5\pi }{12}\right)$

$2+\sqrt{3}$

$\mathrm{tan}\left(-\frac{3\pi }{12}\right)$

$\mathrm{tan}\left(-\frac{3\pi }{8}\right)$

$-1-\sqrt{2}$

For the following exercises, find the exact values of a) $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{x}{2}\right),$ b) $\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{x}{2}\right),$ and c) $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right)$ without solving for $\text{\hspace{0.17em}}x.$

If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x=-\frac{4}{3},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant IV.

If $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x=-\frac{12}{13},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant III.

a) $\text{\hspace{0.17em}}\frac{3\sqrt{13}}{13}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}-\frac{2\sqrt{13}}{13}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}-\frac{3}{2}\text{\hspace{0.17em}}$

If $\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x=7,$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant II.

If $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x=-4,$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant II.

a) $\text{\hspace{0.17em}}\frac{\sqrt{10}}{4}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}\frac{\sqrt{6}}{4}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}\frac{\sqrt{15}}{3}\text{\hspace{0.17em}}$

For the following exercises, use [link] to find the requested half and double angles.

Find $\text{\hspace{0.17em}}\mathrm{sin}\left(2\theta \right),\mathrm{cos}\left(2\theta \right),$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(2\theta \right).$

Find $\text{\hspace{0.17em}}\mathrm{sin}\left(2\alpha \right),\mathrm{cos}\left(2\alpha \right),$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(2\alpha \right).$

$\frac{120}{169},–\frac{119}{169},–\frac{120}{119}$

Find $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\theta }{2}\right),\mathrm{cos}\left(\frac{\theta }{2}\right),$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\theta }{2}\right).$

Find $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha }{2}\right),\mathrm{cos}\left(\frac{\alpha }{2}\right),$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{\alpha }{2}\right).$

$\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13},\frac{2}{3}$

For the following exercises, simplify each expression. Do not evaluate.

${\mathrm{cos}}^{2}\left(28°\right)-{\mathrm{sin}}^{2}\left(28°\right)$

$2{\mathrm{cos}}^{2}\left(37°\right)-1$

$\mathrm{cos}\left(74°\right)$

$1-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\left(17°\right)$

${\mathrm{cos}}^{2}\left(9x\right)-{\mathrm{sin}}^{2}\left(9x\right)$

$\mathrm{cos}\left(18x\right)$

$4\text{\hspace{0.17em}}\mathrm{sin}\left(8x\right)\text{\hspace{0.17em}}\mathrm{cos}\left(8x\right)$

$6\text{\hspace{0.17em}}\mathrm{sin}\left(5x\right)\text{\hspace{0.17em}}\mathrm{cos}\left(5x\right)$

$3\mathrm{sin}\left(10x\right)$

For the following exercises, prove the given identity.

${\left(\mathrm{sin}\text{\hspace{0.17em}}t-\mathrm{cos}\text{\hspace{0.17em}}t\right)}^{2}=1-\mathrm{sin}\left(2t\right)$

$\mathrm{sin}\left(2x\right)=-2\text{\hspace{0.17em}}\mathrm{sin}\left(-x\right)\text{\hspace{0.17em}}\mathrm{cos}\left(-x\right)$

$-2\text{\hspace{0.17em}}\mathrm{sin}\left(-x\right)\mathrm{cos}\left(-x\right)=-2\left(-\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)\right)=\mathrm{sin}\left(2x\right)$

$\mathrm{cot}\text{\hspace{0.17em}}x-\mathrm{tan}\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}\mathrm{cot}\left(2x\right)$

$\frac{\mathrm{sin}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}{\mathrm{tan}}^{2}\theta =\mathrm{tan}\text{\hspace{0.17em}}\theta$

$\begin{array}{ccc}\hfill \frac{\mathrm{sin}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}{\mathrm{tan}}^{2}\theta & =& \frac{2\mathrm{sin}\left(\theta \right)\mathrm{cos}\left(\theta \right)}{1+{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{\mathrm{tan}}^{2}\theta =\hfill \\ \hfill \frac{2\mathrm{sin}\left(\theta \right)\mathrm{cos}\left(\theta \right)}{2{\mathrm{cos}}^{2}\theta }{\mathrm{tan}}^{2}\theta & =& \frac{\mathrm{sin}\left(\theta \right)}{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{tan}}^{2}\theta =\hfill \\ \hfill \mathrm{cot}\left(\theta \right){\mathrm{tan}}^{2}\theta & =& \mathrm{tan}\text{\hspace{0.17em}}\theta \hfill \end{array}$

For the following exercises, rewrite the expression with an exponent no higher than 1.

${\mathrm{cos}}^{2}\left(5x\right)$

${\mathrm{cos}}^{2}\left(6x\right)$

$\frac{1+\mathrm{cos}\left(12x\right)}{2}$

${\mathrm{sin}}^{4}\left(8x\right)$

${\mathrm{sin}}^{4}\left(3x\right)$

$\frac{3+\mathrm{cos}\left(12x\right)-4\mathrm{cos}\left(6x\right)}{8}$

${\mathrm{cos}}^{2}x{\text{\hspace{0.17em}}\mathrm{sin}}^{4}x$

${\mathrm{cos}}^{4}x{\text{\hspace{0.17em}}\mathrm{sin}}^{2}x$

$\frac{2+\mathrm{cos}\left(2x\right)-2\mathrm{cos}\left(4x\right)-\mathrm{cos}\left(6x\right)}{32}$

${\mathrm{tan}}^{2}x{\text{\hspace{0.17em}}\mathrm{sin}}^{2}x$

## Technology

For the following exercises, reduce the equations to powers of one, and then check the answer graphically.

${\mathrm{tan}}^{4}x$

$\frac{3+\mathrm{cos}\left(4x\right)-4\mathrm{cos}\left(2x\right)}{3+\mathrm{cos}\left(4x\right)+4\mathrm{cos}\left(2x\right)}$

${\mathrm{sin}}^{2}\left(2x\right)$

${\mathrm{sin}}^{2}x{\text{\hspace{0.17em}}\mathrm{cos}}^{2}x$

$\frac{1-\mathrm{cos}\left(4x\right)}{8}$

${\mathrm{tan}}^{2}x\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

${\mathrm{tan}}^{4}x{\text{\hspace{0.17em}}\mathrm{cos}}^{2}x$

$\frac{3+\mathrm{cos}\left(4x\right)-4\mathrm{cos}\left(2x\right)}{4\left(\mathrm{cos}\left(2x\right)+1\right)}$

${\mathrm{cos}}^{2}x\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)$

${\mathrm{cos}}^{2}\left(2x\right)\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{\left(1+\mathrm{cos}\left(4x\right)\right)\mathrm{sin}\text{\hspace{0.17em}}x}{2}$

${\mathrm{tan}}^{2}\left(\frac{x}{2}\right)\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

For the following exercises, algebraically find an equivalent function, only in terms of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and/or $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x,$ and then check the answer by graphing both functions.

$\mathrm{sin}\left(4x\right)$

$4\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)$

$\mathrm{cos}\left(4x\right)$

## Extensions

For the following exercises, prove the identities.

$\mathrm{sin}\left(2x\right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{1+{\mathrm{tan}}^{2}x}$

$\begin{array}{l}\frac{2\mathrm{tan}\text{\hspace{0.17em}}x}{1+{\mathrm{tan}}^{2}x}=\frac{\frac{2\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}}{1+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}=\frac{\frac{2\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}}{\frac{{\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}=\\ \frac{2\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}.\frac{{\mathrm{cos}}^{2}x}{1}=2\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x=\mathrm{sin}\left(2x\right)\end{array}$

$\mathrm{cos}\left(2\alpha \right)=\frac{1-{\mathrm{tan}}^{2}\alpha }{1+{\mathrm{tan}}^{2}\alpha }$

$\mathrm{tan}\left(2x\right)=\frac{2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x}{2{\mathrm{cos}}^{2}x-1}$

$\frac{2\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x}{2{\mathrm{cos}}^{2}x-1}=\frac{\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}=\mathrm{tan}\left(2x\right)$

${\left({\mathrm{sin}}^{2}x-1\right)}^{2}=\mathrm{cos}\left(2x\right)+{\mathrm{sin}}^{4}x$

$\mathrm{sin}\left(3x\right)=3\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{3}x$

$\begin{array}{ccc}\hfill \mathrm{sin}\left(x+2x\right)& =& \mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(2x\right)\mathrm{cos}\text{\hspace{0.17em}}x\hfill \\ & =& \mathrm{sin}\text{\hspace{0.17em}}x\left({\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right)+2\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x\hfill \\ & =& \mathrm{sin}\text{\hspace{0.17em}}x{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{3}x+2\mathrm{sin}\text{\hspace{0.17em}}x{\mathrm{cos}}^{2}x\hfill \\ & =& 3\mathrm{sin}\text{\hspace{0.17em}}x{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{3}x\hfill \end{array}$

$\mathrm{cos}\left(3x\right)={\mathrm{cos}}^{3}x-3{\mathrm{sin}}^{2}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{1+\mathrm{cos}\left(2t\right)}{\mathrm{sin}\left(2t\right)-\mathrm{cos}\text{\hspace{0.17em}}t}=\frac{2\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t}{2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t-1}$

$\begin{array}{ccc}\hfill \frac{1+\mathrm{cos}\left(2t\right)}{\mathrm{sin}\left(2t\right)-\mathrm{cos}t}& =& \frac{1+2{\mathrm{cos}}^{2}t-1}{2\mathrm{sin}t\mathrm{cos}t-\mathrm{cos}t}\hfill \\ & =& \frac{2{\mathrm{cos}}^{2}t}{\mathrm{cos}t\left(2\mathrm{sin}t-1\right)}\hfill \\ & =& \frac{2\mathrm{cos}t}{2\mathrm{sin}t-1}\hfill \end{array}$

$\mathrm{sin}\left(16x\right)=16\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)\mathrm{cos}\left(4x\right)\mathrm{cos}\left(8x\right)$

$\mathrm{cos}\left(16x\right)=\left({\mathrm{cos}}^{2}\left(4x\right)-{\mathrm{sin}}^{2}\left(4x\right)-\mathrm{sin}\left(8x\right)\right)\left({\mathrm{cos}}^{2}\left(4x\right)-{\mathrm{sin}}^{2}\left(4x\right)+\mathrm{sin}\left(8x\right)\right)$

$\begin{array}{ccc}\hfill \left({\mathrm{cos}}^{2}\left(4x\right)-{\mathrm{sin}}^{2}\left(4x\right)-\mathrm{sin}\left(8x\right)\right)\left({\mathrm{cos}}^{2}\left(4x\right)-{\mathrm{sin}}^{2}\left(4x\right)+\mathrm{sin}\left(8x\right)\right)& =& \\ & =& \left(\mathrm{cos}\left(8x\right)-\mathrm{sin}\left(8x\right)\right)\left(\mathrm{cos}\left(8x\right)+\mathrm{sin}\left(8x\right)\right)\hfill \\ & =& {\mathrm{cos}}^{2}\left(8x\right)-{\mathrm{sin}}^{2}\left(8x\right)\hfill \\ & =& \mathrm{cos}\left(16x\right)\hfill \end{array}$

#### Questions & Answers

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
Kaitlyn Reply
The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
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Aaron Reply

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