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Question

Question: Write \(\cos 20^\circ \) in terms of \(\sin \)?...

Write cos20\cos 20^\circ in terms of sin\sin ?

Explanation

Solution

Hint : Here, in the given question we need to write cos20\cos 20^\circ in terms of sine\sin e. cosine\cos ine and sine\sin e are cofunctions. Cofunction is a trigonometric function whose value for the complement of an angle is equal to the value of a given trigonometric function of the angle itself. So, as we know that sine\sin e and cosine\cos ine are complementary angles. Two angles are said to be complementary, if their sum is 9090^\circ . For example: θ\theta and (90θ)\left( {90^\circ - \theta } \right) are complementary angles for an acute angle θ\theta because their sum is 9090^\circ . Similarly sine\sin e and cosine\cos ine are complementary angles. Therefore, cos\cos of an angle = sin\sin of its complementary angle. We can write it as: cosθ=sin(90θ)\cos \theta = \sin \left( {90^\circ - \theta } \right). So, we will write cos20\cos 20^\circ as sin(9020)\sin \left( {90^\circ - 20^\circ } \right), and proceed.

Complete step-by-step answer :
Given, cos20\cos 20^\circ
As we know cosθ=sin(90θ)\cos \theta = \sin \left( {90^\circ - \theta } \right). Therefore, the above written function can be written in terms of sine\sin e as:
cos20=sin(9020)\Rightarrow \cos 20^\circ = \sin \left( {90^\circ - 20^\circ } \right)
On subtracting the terms inside the bracket, we get
cos20=sin(70)\Rightarrow \cos 20^\circ = \sin \left( {70^\circ } \right)
Hence, the value of cos20\cos 20^\circ in terms of sine\sin e is sin70\sin 70^\circ .
So, the correct answer is “sin70\sin 70^\circ ”.

Note : Remember that conjunctions are pairs of trigonometric functions. Remember a simple relationship: sin(α)=cos(β)\sin \left( \alpha \right) = \cos \left( \beta \right) where α+β=90\alpha + \beta = 90^\circ . This also holds true for cosec\cos ec and sec\sec , and tan\tan and cot\cot . Here, in the given question, we wrote cos20\cos 20^\circ in terms of sine\sin e, as you can see cos20\cos 20^\circ is a very small angle that is why we used cosθ=sin(90θ)\cos \theta = \sin \left( {90^\circ - \theta } \right) formula, as know (90θ)\left( {90^\circ - \theta } \right) lies in the first quadrant and in first quadrant all the trigonometric ratios are positive. While writing one trigonometric ratio in terms of another trigonometric ratio, one must remember in which quadrant sign is positive or negative for that particular trigonometric function.