
Exact value of \[\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ \] is:
A). \[1\]
B). \[\dfrac{1}{{\sqrt 2 }}\]
C). \[\sqrt 2 \]
D). zero
Answer
474.3k+ views
Hint: In the given question, we have been given an expression involving the use of trigonometric functions. The angles are not the ones given in the range of the standard table. We are going to solve it by converting the trigonometric functions into their primitive form. Then we are going to convert the trigonometric functions into such forms that their angles are equal using the appropriate formulae and then solve to get the answer.
Formula used:
We are going to use the formula:
\[2{\sin ^2}\theta = 1 - \cos \left( {2\theta } \right)\]
Complete step by step solution:
The given expression is \[\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ \].
Applying the formula of \[2{\sin ^2}\theta = 1 - \cos \left( {2\theta } \right)\], we have,
\[2{\sin ^2}55^\circ = 1 - \cos 110^\circ \]
So, we have,
\[\cos 20^\circ + 1 - \cos 110^\circ - \sqrt 2 \sin 65^\circ \]
Now, \[\sin \left( \theta \right) = \cos \left( {90 - \theta } \right)\]
So, we have,
\[\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \sin 65^\circ \]
Now, \[\sin 65^\circ = \sin \left( {20 + 45} \right)^\circ \], and using the sum formula,
\[\cos 20^\circ + 1 + sin20^\circ - \sqrt 2 \left( {\sin 45^\circ \times \cos 20^\circ + \cos 45^\circ \times \sin 20^\circ } \right)\]
Putting in the values,
\[\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} \times \cos 20^\circ + \dfrac{1}{{\sqrt 2 }} \times \sin 20^\circ } \right)\]
Opening the bracket,
\[\cos 20^\circ + \sin 20^\circ + 1 - \cos 20^\circ - \sin 20^\circ \]
Hence, the expression is equal to \[1\].
Thus, the correct option is A.
Note: In this question, we had to find the sum of given trigonometric functions. We solved this question by converting the functions into their primitive form. Then we applied the appropriate identities, used their result to get to the point where the angles of functions were equal. We have to remember that when there is no apparent identity that we can apply, we have to think of some straight-forward answer, involving the use of the basic knowledge of the subjects’ properties.
Formula used:
We are going to use the formula:
\[2{\sin ^2}\theta = 1 - \cos \left( {2\theta } \right)\]
Complete step by step solution:
The given expression is \[\cos 20^\circ + 2{\sin ^2}55^\circ - \sqrt 2 \sin 65^\circ \].
Applying the formula of \[2{\sin ^2}\theta = 1 - \cos \left( {2\theta } \right)\], we have,
\[2{\sin ^2}55^\circ = 1 - \cos 110^\circ \]
So, we have,
\[\cos 20^\circ + 1 - \cos 110^\circ - \sqrt 2 \sin 65^\circ \]
Now, \[\sin \left( \theta \right) = \cos \left( {90 - \theta } \right)\]
So, we have,
\[\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \sin 65^\circ \]
Now, \[\sin 65^\circ = \sin \left( {20 + 45} \right)^\circ \], and using the sum formula,
\[\cos 20^\circ + 1 + sin20^\circ - \sqrt 2 \left( {\sin 45^\circ \times \cos 20^\circ + \cos 45^\circ \times \sin 20^\circ } \right)\]
Putting in the values,
\[\cos 20^\circ + 1 + \sin 20^\circ - \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} \times \cos 20^\circ + \dfrac{1}{{\sqrt 2 }} \times \sin 20^\circ } \right)\]
Opening the bracket,
\[\cos 20^\circ + \sin 20^\circ + 1 - \cos 20^\circ - \sin 20^\circ \]
Hence, the expression is equal to \[1\].
Thus, the correct option is A.
Note: In this question, we had to find the sum of given trigonometric functions. We solved this question by converting the functions into their primitive form. Then we applied the appropriate identities, used their result to get to the point where the angles of functions were equal. We have to remember that when there is no apparent identity that we can apply, we have to think of some straight-forward answer, involving the use of the basic knowledge of the subjects’ properties.
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