COS A PLUS B FORMULA

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cos (A + B) = cos A cos B - sin A sin B

cos (A - B) = cos A cos B + sin A sin B 

Express each of the following in the form sin ฮฑ, where ฮฑ is acute.

Problem 1 :

cos 14ยบ cos 39ยบ - sin 14ยบ sin 39ยบ

Solution :

cos (A + B) = cos A cos B - sin A sin B 

= cos 14ยบ cos 39ยบ - sin 14ยบ sin 39ยบ

= cos (14ยบ + 39ยบ)

= cos 53ยบ

Problem 2 :

Express as a single trigonometric ratio.

a. cos A cos 2A - sin A sin 2A

b. cos A cos 3A + sin A sin 3A

Solution :

a. 

cos (A + B) = cos A cos B - sin A sin B 

= cos A cos 2A - sin A sin 2A

= cos (A + 2A)

=  cos 3A

b.

cos (A - B) = cos A cos B + sin A sin B 

= cos A cos 3A + sin A sin 3A

= cos (A + 3A)

=  cos 4A

Problem 3 :

cos 38ยบ cos 8ยบ + sin 38ยบ sin 8ยบ is equal to.

a)  cos 30ยบ  b)  cos 60ยบ   c)   cos 45ยบ   d)  cos 38ยบ

Solution  :

cos (A - B) = cos A cos B + sin A sin B 

= cos 38ยบ cos 8ยบ + sin 38ยบ sin 8ยบ

= cos (38ยบ - 8ยบ)

= cos 30ยบ

So, option a) is correct.

Problem 4 :

Find the value of k such that for all real values of x.

cos x + ๐œ‹3 - cos x - ๐œ‹3 โ‰ก k sin x

Solution :

Given, cos x + ๐œ‹3 - cos x - ๐œ‹3 โ‰ก k sin xcos c - cos d = -2 sin c + d2ยท sin c - d2 -2 sin x + ๐œ‹3 + x - ๐œ‹32 ยท sin x + ๐œ‹3 - x - ๐œ‹32 = k sin x -2 sin x + ๐œ‹3 + x - ๐œ‹32 ยท sin x + ๐œ‹3 - x + ๐œ‹32 = k sin x -2 sin 2x 2 ยท sin 2๐œ‹32 = k sin x -2 sin 2x 2 ยท sin 2๐œ‹3 ร— 12 = k sin x-2 sin x ยท sin 2๐œ‹6 = k sin x-2 sin x ยท sin ๐œ‹3 = k sin x-2 sin x ยท32= k sin x- 3 sin x = k sin xk = - 3 sin xsin xk = -3

So, the value of k is -โˆš3.

Problem 5 :

Prove each identity.

cos x - cos x - ๐œ‹3 โ‰ก cos x + ๐œ‹3

Solution :

Given, cos x - cos x - ๐œ‹3 โ‰ก cos x + ๐œ‹3cos x = cos x + ๐œ‹3 + cos x - ๐œ‹3cos (a + b) = cos a cos b - sin a sin bcos (a - b) = cos a cos b + sin a sin bcos x = cos x cos ๐œ‹3 - sin x sin ๐œ‹3 +cos x cos ๐œ‹3 + sin x sin ๐œ‹3cos x = 2 cos x cos ๐œ‹3 cos x = 2 cos x 12cos x = cos x

Problem 6 :

Rewrite each expression as a single trigonometric ratio.

cos 4x cos 3x - sin 4x sin 3x

Solution :

cos (A + B) = cos A cos B - sin A sin B 

= cos 4x cos 3x - sin 4x sin 3x

= cos (4x + 3x)

= cos 7x

Problem 7 :

Rewrite each expression as a single trigonometric ratio, and then evaluate the ratio.

cos 5๐œ‹12 cos ๐œ‹12 + sin 5๐œ‹12 sin ๐œ‹12

Solution :

cos (A - B) = cos A cos B + sin A sin B= cos 5๐œ‹12 cos ๐œ‹12 + sin 5๐œ‹12 sin ๐œ‹12= cos 5๐œ‹12 - ๐œ‹12=cos 4๐œ‹12 = cos ๐œ‹3= 12

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