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Question: Which of the following is incorrect? A. \[{\cos ^4}\theta - {\sin ^4}\theta = {\cos ^2}\theta - {\...

Which of the following is incorrect?
A. cos⁡4θ−sin⁡4θ=cos⁡2θ−sin⁡2θ{\cos ^4}\theta - {\sin ^4}\theta = {\cos ^2}\theta - {\sin ^2}\theta
B. 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta
C. sin⁡40∘+cos⁡50∘=2sin⁡40∘\sin {40^ \circ } + \cos {50^ \circ } = 2\sin {40^ \circ }
D. sin⁡2θ+cos⁡2(−θ)=−1{\sin ^2}\theta + {\cos ^2}\left( { - \theta } \right) = - 1

Explanation

Solution

Given are the trigonometric ratios there are various standard identities also. We will check them one by one. We have to find the incorrect statement. We have to use the sum and difference formula as well if required.

Complete step by step answer:
A. cos⁡4θ−sin⁡4θ=cos⁡2θ−sin⁡2θ{\cos ^4}\theta - {\sin ^4}\theta = {\cos ^2}\theta - {\sin ^2}\theta
In this statement we will modify the LHS
(cos⁡2θ)2−(sin⁡2θ)2=cos⁡2θ−sin⁡2θ{\left( {{{\cos }^2}\theta } \right)^2} - {\left( {{{\sin }^2}\theta } \right)^2} = {\cos ^2}\theta - {\sin ^2}\theta
Now we will apply the identities on LHS as,
(cos⁡2θ+sin⁡2θ)(cos⁡2θ−sin⁡2θ)=cos⁡2θ−sin⁡2θ\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = {\cos ^2}\theta - {\sin ^2}\theta
Now we can cancel the common terms from both the sides,
(cos⁡2θ+sin⁡2θ)=1\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) = 1
This is the standard identity. Thus this is the correct statement.

B. 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta
This is the standard identity of trigonometric ratios and need not to be verified.This is the correct statement.

C. sin⁡40∘+cos⁡50∘=2sin⁡40∘\sin {40^ \circ } + \cos {50^ \circ } = 2\sin {40^ \circ }
This can be solved using negative angles.
We know that, cos⁡(90∘−θ)=sin⁡θ\cos \left( {{{90}^ \circ } - \theta } \right) = \sin \theta
Thus we can write,
sin⁡40∘+cos⁡50∘=sin⁡40∘+cos⁡(90∘−40∘)\sin {40^ \circ } + \cos {50^ \circ } = \sin {40^ \circ } + \cos \left( {{{90}^ \circ } - {{40}^ \circ }} \right)
⇒sin⁡40∘+cos⁡50∘=sin⁡40∘+sin⁡40∘\Rightarrow \sin {40^ \circ } + \cos {50^ \circ }= \sin {40^ \circ } + \sin {40^ \circ }
On adding we get,
sin⁡40∘+cos⁡50∘=2sin⁡40∘\sin {40^ \circ } + \cos {50^ \circ }= 2\sin {40^ \circ }
Thus this is also a correct statement.

D. sin⁡2θ+cos⁡2(−θ)=−1{\sin ^2}\theta + {\cos ^2}\left( { - \theta } \right) = - 1
We can express the LHS as,
sin⁡2θ+(cos⁡(−θ))2{\sin ^2}\theta + {\left( {\cos \left( { - \theta } \right)} \right)^2}
We know that, cos⁡(−θ)=−cos⁡θ\cos \left( { - \theta } \right) = - \cos \theta
sin⁡2θ+(−cos⁡θ)2{\sin ^2}\theta + {\left( { - \cos \theta } \right)^2}
Square of any negative term is always positive.
sin⁡2θ+cos⁡2θ{\sin ^2}\theta + {\cos ^2}\theta
We know this standard identity equals to 1.
11
But the question says it is minus 1. Thus this is an incorrect identity.

Thus option D is the answer.

Note: Here note that even a slight minus sign can change the whole problem. So be aware of that. Also note that, when we solve these types of problems we might deal with the double angle and triple angle formulae. So when we use other identities by replacing the original we should use standard identities. Sometimes a combination of algebraic and trigonometric identities is required. In the statements above we used this in the first statement.Along with this we also need to know the negative angles of the standard ratios.Also note that, when we solve this type of problems it is not necessary that every time the RHS should be equal to LHS. We need not to stretch the answer unnecessarily. Just go with the correct and to the point steps.