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                     cos ( α β ) = cos α cos β + sin α sin β                   cos ( α + β ) = ( cos α cos β sin α sin β ) ____________________________________________________ cos ( α β ) cos ( α + β ) = 2 sin α sin β

Then, we divide by 2 to isolate the product of sines:

sin α sin β = 1 2 [ cos ( α β ) cos ( α + β ) ]

Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas.

The product-to-sum formulas

The product-to-sum formulas are as follows:

cos α cos β = 1 2 [ cos ( α β ) + cos ( α + β ) ]
sin α cos β = 1 2 [ sin ( α + β ) + sin ( α β ) ]
sin α sin β = 1 2 [ cos ( α β ) cos ( α + β ) ]
cos α sin β = 1 2 [ sin ( α + β ) sin ( α β ) ]

Express the product as a sum or difference

Write cos ( 3 θ ) cos ( 5 θ ) as a sum or difference.

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.

cos α cos β = 1 2 [ cos ( α β ) + cos ( α + β ) ] cos ( 3 θ ) cos ( 5 θ ) = 1 2 [ cos ( 3 θ 5 θ ) + cos ( 3 θ + 5 θ ) ] = 1 2 [ cos ( 2 θ ) + cos ( 8 θ ) ] Use even-odd identity .
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Use the product-to-sum formula to evaluate cos 11 π 12 cos π 12 .

2 3 4

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Expressing sums as products

Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine . Let u + v 2 = α and u v 2 = β .

Then,

α + β = u + v 2 + u v 2 = 2 u 2 = u α β = u + v 2 u v 2 = 2 v 2 = v

Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have

sin α cos β = 1 2 [ sin ( α + β ) + sin ( α β ) ] sin ( u + v 2 ) cos ( u v 2 ) = 1 2 [ sin u + sin v ] Substitute for ( α + β )  and  ( α β ) 2 sin ( u + v 2 ) cos ( u v 2 ) = sin u + sin v

The other sum-to-product identities are derived similarly.

Sum-to-product formulas

The sum-to-product formulas are as follows:

sin α + sin β = 2 sin ( α + β 2 ) cos ( α β 2 )
sin α sin β = 2 sin ( α β 2 ) cos ( α + β 2 )
cos α cos β = −2 sin ( α + β 2 ) sin ( α β 2 )
cos α + cos β = 2 cos ( α + β 2 ) cos ( α β 2 )

Writing the difference of sines as a product

Write the following difference of sines expression as a product: sin ( 4 θ ) sin ( 2 θ ) .

We begin by writing the formula for the difference of sines.

sin α sin β = 2 sin ( α β 2 ) cos ( α + β 2 )

Substitute the values into the formula, and simplify.

sin ( 4 θ ) sin ( 2 θ ) = 2 sin ( 4 θ 2 θ 2 ) cos ( 4 θ + 2 θ 2 ) = 2 sin ( 2 θ 2 ) cos ( 6 θ 2 ) = 2 sin θ cos ( 3 θ )
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Use the sum-to-product formula to write the sum as a product: sin ( 3 θ ) + sin ( θ ) .

2 sin ( 2 θ ) cos ( θ )

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Evaluating using the sum-to-product formula

Evaluate cos ( 15° ) cos ( 75° ) . Check the answer with a graphing calculator.

We begin by writing the formula for the difference of cosines.

cos α cos β = 2 sin ( α + β 2 ) sin ( α β 2 )

Then we substitute the given angles and simplify.

cos ( 15° ) cos ( 75° ) = −2 sin ( 15° + 75° 2 ) sin ( 15° 75° 2 ) = −2 sin ( 45° ) sin ( −30° ) = −2 ( 2 2 ) ( 1 2 ) = 2 2
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Proving an identity

Prove the identity:

cos ( 4 t ) cos ( 2 t ) sin ( 4 t ) + sin ( 2 t ) = tan t

We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.

cos ( 4 t ) cos ( 2 t ) sin ( 4 t ) + sin ( 2 t ) = 2 sin ( 4 t + 2 t 2 ) sin ( 4 t 2 t 2 ) 2 sin ( 4 t + 2 t 2 ) cos ( 4 t 2 t 2 ) = 2 sin ( 3 t ) sin t 2 sin ( 3 t ) cos t = 2 sin ( 3 t ) sin t 2 sin ( 3 t ) cos t = sin t cos t = tan t
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Questions & Answers

what is the answer to dividing negative index
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In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
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tanh`(x-iy) =A+iB, find A and B
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B=Ai-itan(hx-hiy)
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If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
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exponential series
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Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
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Practice Key Terms 2

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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