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Question: Find the value of \(\cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ }\)...

Find the value of cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ }

Explanation

Solution

This is a trigonometric problem which includes one of the trigonometric sum and difference formulas of sine and cosine. While solving these kinds of problems the trigonometric identities and trigonometric formulas are very important and these formulas and identities are to be remembered. Here the trigonometric difference formula is used which is :
⇒cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\Rightarrow \cos (A - B) = \cos A\cos B + \sin A\sin B

Complete step-by-step solution:
Here consider the given problem cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ },
Compare the given problem with the trigonometric difference formula which is:
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B
Here A=70∘A = {70^ \circ } and B=10∘B = {10^ \circ }
Applying the formula to the problem:
⇒cos⁡(70∘−10∘)=cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\Rightarrow \cos ({70^ \circ } - {10^ \circ }) = \cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ }
⇒cos⁡(60∘)=cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\Rightarrow \cos ({60^ \circ }) = \cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ }
Re-writing the equation which is given below:
⇒cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘=cos⁡(60∘)\Rightarrow \cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ } = \cos ({60^ \circ })
We know that cos⁡(60∘)=12\cos ({60^ \circ }) = \dfrac{1}{2}
⇒cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘=12\Rightarrow \cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ } = \dfrac{1}{2}

The value of cos⁡70∘cos⁡10∘+sin⁡70∘sin⁡10∘\cos {70^ \circ }\cos {10^ \circ } + \sin {70^ \circ }\sin {10^ \circ } is 12\dfrac{1}{2}

Note: The most important thing to remember is that to not to confuse between the sum and difference trigonometric formulas of sine and cosine, which is: cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos (A \pm B) = \cos A\cos B \mp \sin A\sin B but whereas for sine it is: sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B