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Find the value of cos 28°​cos 62°​– sin 28°​sin 62°​

Sagot :

stigawithfun

Answer:

cos 28°​cos 62°​– sin 28°​sin 62°​ = 0

Step-by-step explanation:

From one of the trigonometric identities stated as follows;

cos(A+B) = cosAcosB - sinAsinB             -----------------(i)

We can apply such identity to solve the given expression.

Given:

cos 28°​cos 62°​– sin 28°​sin 62°​

Comparing the given expression with the right hand side of equation (i), we see that;

A = 28°

B = 62°

∴ Substitute these values into equation (i) to have;

⇒ cos(28°+62°) = cos28°cos62° - sin28°sin62°

Solve the left hand side.

⇒ cos(90°) = cos28°cos62° - sin28°sin62°

⇒ 0 = cos28°cos62° - sin28°sin62°     (since cos 90° = 0)

Therefore,

cos28°cos62° - sin28°sin62° = 0

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