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Question: What is the value of \(({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ })?\)...

What is the value of (cos⁡267∘ - sin⁡223∘)?({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ })?

Explanation

Solution

Hint: Here convert cos⁡θ\cos \theta into sinθsin \theta by using the trigonometric relation cos⁡2(90∘ - θ) = sin⁡2θ{\cos ^2}({90^ \circ }{\text{ - }}\theta ){\text{ = }}{\sin ^2}\theta .

Complete step by step solution:

Given equation is (cos⁡267∘ - sin⁡223∘)({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ })

Now, we can write this equation in this form i.e.

(cos⁡267∘ - sin⁡223∘)=cos⁡2(90∘ - 23∘) - sin⁡223∘({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ }) = {\cos ^2}({90^ \circ }{\text{ - 2}}{{\text{3}}^ \circ }{\text{) - }}{\sin ^2}{23^ \circ } ......................................(1)

We know that

cos⁡2(90∘ - θ) = sin⁡2θ{\cos ^2}({90^ \circ }{\text{ - }}\theta ){\text{ = }}{\sin ^2}\theta

Now using this concept we can write the equation (1) in this form i.e.

(cos⁡267∘ - sin⁡223∘) = sin⁡223∘ - sin⁡223∘({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ }){\text{ = }}{\sin ^2}{23^ \circ }{\text{ - }}{\sin ^2}{23^ \circ }

(cos⁡267∘ - sin⁡223∘)= 0({\cos ^2}{67^ \circ }{\text{ - }}{\sin ^2}{23^ \circ }) = {\text{ 0}}

Note: - These types of problems can be solved by converting either cos⁡θ\cos \theta into sin⁡θ\sin \theta or sinθsin \theta into cos⁡θ\cos \theta. Here we have converted cos⁡θ\cos \theta into sin⁡θ\sin \theta in the similar way we can do it for other questions as well.