# 9.3 Double-angle, half-angle, and reduction formulas  (Page 4/8)

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Given that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =-\frac{4}{5}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ lies in quadrant IV, find the exact value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\left(\frac{\alpha }{2}\right).$

$-\frac{2}{\sqrt{5}}$

## Finding the measurement of a half angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}\text{\hspace{0.17em}}$ for higher-level competition, what is the measurement of the angle for novice competition?

Since the angle for novice competition measures half the steepness of the angle for the high level competition, and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}\text{\hspace{0.17em}}$ for high competition, we can find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See [link] .

$\begin{array}{ccc}\hfill {3}^{2}+{5}^{2}& =& 34\hfill \\ \hfill c& =& \sqrt{34}\hfill \end{array}$

We see that $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}.\text{\hspace{0.17em}}$ We can use the half-angle formula for tangent: $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}=\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{cos}\text{\hspace{0.17em}}\theta }}.\text{\hspace{0.17em}}$ Since $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the first quadrant, so is $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}.\text{\hspace{0.17em}}$

$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\theta }{2}& =& \sqrt{\frac{1-\frac{3\sqrt{34}}{34}}{1+\frac{3\sqrt{34}}{34}}}\hfill \\ & =& \sqrt{\frac{\frac{34-3\sqrt{34}}{34}}{\frac{34+3\sqrt{34}}{34}}}\hfill \\ & =& \sqrt{\frac{34-3\sqrt{34}}{34+3\sqrt{34}}}\hfill \\ & \approx & 0.57\hfill \end{array}$

We can take the inverse tangent to find the angle: $\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(0.57\right)\approx 29.7°.\text{\hspace{0.17em}}$ So the angle of the ramp for novice competition is $\text{\hspace{0.17em}}\approx 29.7°.$

Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.

## Key equations

 Double-angle formulas $\begin{array}{ccc}\hfill \mathrm{sin}\left(2\theta \right)& =& 2\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ \hfill \mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ & =& 1-2{\mathrm{sin}}^{2}\theta \hfill \\ & =& 2{\mathrm{cos}}^{2}\theta -1\hfill \\ \hfill \mathrm{tan}\left(2\theta \right)& =& \frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\hfill \end{array}$ Reduction formulas $\begin{array}{ccc}\hfill {\mathrm{sin}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{cos}}^{2}\theta & =& \frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ \hfill {\mathrm{tan}}^{2}\theta & =& \frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}\hfill \end{array}$ Half-angle formulas $\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \hfill \mathrm{tan}\text{\hspace{0.17em}}\frac{\alpha }{2}& =& ±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ & =& \frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$

## Key concepts

• Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See [link] , [link] , [link] , and [link] .
• Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See [link] and [link] .
• Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See [link] , [link] , and [link] .

## Verbal

Explain how to determine the reduction identities from the double-angle identity $\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x.$

Use the Pythagorean identities and isolate the squared term.

Explain how to determine the double-angle formula for $\text{\hspace{0.17em}}\mathrm{tan}\left(2x\right)\text{\hspace{0.17em}}$ using the double-angle formulas for $\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right).$

We can determine the half-angle formula for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right)=\frac{\sqrt{1-\mathrm{cos}\text{\hspace{0.17em}}x}}{\sqrt{1+\mathrm{cos}\text{\hspace{0.17em}}x}}\text{\hspace{0.17em}}$ by dividing the formula for $\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{x}{2}\right)\text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{x}{2}\right).\text{\hspace{0.17em}}$ Explain how to determine two formulas for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right)\text{\hspace{0.17em}}$ that do not involve any square roots.

$\text{\hspace{0.17em}}\frac{1-\mathrm{cos}\text{\hspace{0.17em}}x}{\mathrm{sin}\text{\hspace{0.17em}}x},\frac{\mathrm{sin}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x},$ multiplying the top and bottom by $\text{\hspace{0.17em}}\sqrt{1-\mathrm{cos}\text{\hspace{0.17em}}x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\sqrt{1+\mathrm{cos}\text{\hspace{0.17em}}x},$ respectively.

For the half-angle formula given in the previous exercise for $\text{\hspace{0.17em}}\mathrm{tan}\left(\frac{x}{2}\right),$ explain why dividing by 0 is not a concern. (Hint: examine the values of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ necessary for the denominator to be 0.)

## Algebraic

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

If $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x=\frac{1}{8},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant I.

a) $\text{\hspace{0.17em}}\frac{3\sqrt{7}}{32}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}\frac{31}{32}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}\frac{3\sqrt{7}}{31}$

If $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x=\frac{2}{3},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant I.

If $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x=-\frac{1}{2},$ and $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in quadrant III.

a) $\text{\hspace{0.17em}}\frac{\sqrt{3}}{2}\text{\hspace{0.17em}}$ b) $\text{\hspace{0.17em}}-\frac{1}{2}\text{\hspace{0.17em}}$ c) $\text{\hspace{0.17em}}-\sqrt{3}\text{\hspace{0.17em}}$

#### Questions & Answers

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
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Damian
I need to learn this trigonometry from A level.. can anyone help here?
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AUSTINE
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cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
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x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
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I mean x²=2x+8 by factorization method
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I think x=-2 or x=4
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x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
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