Question

The two legs of a right triangle are $\sin \theta +\sin \left( \dfrac{3\pi }{2}-\theta \right)$ and $\cos \theta -\cos \left( \dfrac{3\pi }{2}-\theta \right)$. The length of its hypotenuse is:
A. $1$
B. $\sqrt{2}$
C. $2$
D. Some function of $\theta $

Answer 1

Hint: The given problem is related to the value of sine and cosine of an angle in the third quadrant. Sine and cosine function are negative in the third quadrant. Use this property to find the lengths of the legs of the right triangle. Then use the Pythagoras theorem to determine the length of its hypotenuse.

Complete step-by-step answer:
We know, any angle in the third quadrant is of the form $\left( \dfrac{3\pi }{2}-\theta \right)$ . We also know that sine and cosine functions are negative in the third quadrant. So, the value of $\sin \left( \dfrac{3\pi }{2}-\theta \right)$ will be $-\cos \theta $ and the value of $\cos \left( \dfrac{3\pi }{2}-\theta \right)$ will be $-\sin \theta $ . Now, the length of the legs of the right triangle are given as $\sin \theta +\sin \left( \dfrac{3\pi }{2}-\theta \right)$ and $\cos \theta -\cos \left( \dfrac{3\pi }{2}-\theta \right)$ . But we know that the value of $\sin \left( \dfrac{3\pi }{2}-\theta \right)$ is $-\cos \theta $ and the value of $\cos \left( \dfrac{3\pi }{2}-\theta \right)$ is $-\sin \theta $ . So, the length of the legs of the right triangle are $\sin \theta -\cos \theta $ and $\cos \theta +\sin \theta $ .
Now, we need to find the length of its hypotenuse. We know, the Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the perpendicular sides. So, if $a,b$ and $c$ are the lengths of sides of a right triangle such that $c>b,a$ , then according to the Pythagoras theorem, ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ .
Now, in the given right triangle, the length of the legs are $\sin \theta -\cos \theta $ and $\cos \theta +\sin \theta $ . Let $h$ be the length of the hypotenuse. So, according to the Pythagoras theorem, \[{{h}^{2}}={{\left( \sin \theta -\cos \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}\].
$\Rightarrow {{h}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta +{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\sin \theta \cos \theta $
Now, we know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . So, ${{h}^{2}}=1+1=2$ .
$\Rightarrow h=\sqrt{2}$
Hence, the length of the hypotenuse is $\sqrt{2}$ . Hence, option B. is the correct option.

Note: Some students get confused and write ${{\sin }^{2}}\theta -{{\cos }^{2}}\theta =1$ instead of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Such mistakes should be avoided as it can result in getting wrong answers.