
Write $\cos 20^\circ $ in terms of $\sin $?
Answer
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Hint: Here, in the given question we need to write $\cos 20^\circ $ in terms of $\sin e$. $\cos ine$ and $\sin e$ 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 $\sin e$ and $\cos ine$ are complementary angles. Two angles are said to be complementary, if their sum is $90^\circ $. For example: $\theta $ and $\left( {90^\circ - \theta } \right)$ are complementary angles for an acute angle $\theta $ because their sum is $90^\circ $. Similarly $\sin e$ and $\cos ine$ are complementary angles. Therefore, $\cos $ of an angle = $\sin $ of its complementary angle. We can write it as: $\cos \theta = \sin \left( {90^\circ - \theta } \right)$. So, we will write $\cos 20^\circ $ as $\sin \left( {90^\circ - 20^\circ } \right)$, and proceed.
Complete step-by-step answer:
Given, $\cos 20^\circ $
As we know $\cos \theta = \sin \left( {90^\circ - \theta } \right)$. Therefore, the above written function can be written in terms of $\sin e$ as:
$ \Rightarrow \cos 20^\circ = \sin \left( {90^\circ - 20^\circ } \right)$
On subtracting the terms inside the bracket, we get
$ \Rightarrow \cos 20^\circ = \sin \left( {70^\circ } \right)$
Hence, the value of $\cos 20^\circ $ in terms of $\sin e$ is $\sin 70^\circ $.
So, the correct answer is “$\sin 70^\circ $”.
Note: Remember that conjunctions are pairs of trigonometric functions. Remember a simple relationship: $\sin \left( \alpha \right) = \cos \left( \beta \right)$ where $\alpha + \beta = 90^\circ $. This also holds true for $\cos ec$ and $\sec $, and $\tan $ and $\cot $. Here, in the given question, we wrote $\cos 20^\circ $ in terms of $\sin e$, as you can see $\cos 20^\circ $ is a very small angle that is why we used $\cos \theta = \sin \left( {90^\circ - \theta } \right)$ formula, as know $\left( {90^\circ - \theta } \right)$ 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.
Complete step-by-step answer:
Given, $\cos 20^\circ $
As we know $\cos \theta = \sin \left( {90^\circ - \theta } \right)$. Therefore, the above written function can be written in terms of $\sin e$ as:
$ \Rightarrow \cos 20^\circ = \sin \left( {90^\circ - 20^\circ } \right)$
On subtracting the terms inside the bracket, we get
$ \Rightarrow \cos 20^\circ = \sin \left( {70^\circ } \right)$
Hence, the value of $\cos 20^\circ $ in terms of $\sin e$ is $\sin 70^\circ $.
So, the correct answer is “$\sin 70^\circ $”.
Note: Remember that conjunctions are pairs of trigonometric functions. Remember a simple relationship: $\sin \left( \alpha \right) = \cos \left( \beta \right)$ where $\alpha + \beta = 90^\circ $. This also holds true for $\cos ec$ and $\sec $, and $\tan $ and $\cot $. Here, in the given question, we wrote $\cos 20^\circ $ in terms of $\sin e$, as you can see $\cos 20^\circ $ is a very small angle that is why we used $\cos \theta = \sin \left( {90^\circ - \theta } \right)$ formula, as know $\left( {90^\circ - \theta } \right)$ 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.
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