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Question: The two legs of a right triangle are \(\sin \theta +\sin \left( \dfrac{3\pi }{2}-\theta \right)\) an...

The two legs of a right triangle are sin⁡θ+sin⁡(3π2−θ)\sin \theta +\sin \left( \dfrac{3\pi }{2}-\theta \right) and cos⁡θ−cos⁡(3π2−θ)\cos \theta -\cos \left( \dfrac{3\pi }{2}-\theta \right). The length of its hypotenuse is:
A. 11
B. 2\sqrt{2}
C. 22
D. Some function of θ\theta

Explanation

Solution

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 (3π2−θ)\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⁡(3π2−θ)\sin \left( \dfrac{3\pi }{2}-\theta \right) will be −cos⁡θ-\cos \theta and the value of cos⁡(3π2−θ)\cos \left( \dfrac{3\pi }{2}-\theta \right) will be −sin⁡θ-\sin \theta . Now, the length of the legs of the right triangle are given as sin⁡θ+sin⁡(3π2−θ)\sin \theta +\sin \left( \dfrac{3\pi }{2}-\theta \right) and cos⁡θ−cos⁡(3π2−θ)\cos \theta -\cos \left( \dfrac{3\pi }{2}-\theta \right) . But we know that the value of sin⁡(3π2−θ)\sin \left( \dfrac{3\pi }{2}-\theta \right) is −cos⁡θ-\cos \theta and the value of cos⁡(3π2−θ)\cos \left( \dfrac{3\pi }{2}-\theta \right) is −sin⁡θ-\sin \theta . So, the length of the legs of the right triangle are sin⁡θ−cos⁡θ\sin \theta -\cos \theta and cos⁡θ+sin⁡θ\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,ba,b and cc are the lengths of sides of a right triangle such that c>b,ac>b,a , then according to the Pythagoras theorem, c2=a2+b2{{c}^{2}}={{a}^{2}}+{{b}^{2}} .
Now, in the given right triangle, the length of the legs are sin⁡θ−cos⁡θ\sin \theta -\cos \theta and cos⁡θ+sin⁡θ\cos \theta +\sin \theta . Let hh be the length of the hypotenuse. So, according to the Pythagoras theorem, h2=(sin⁡θ−cos⁡θ)2+(cos⁡θ+sin⁡θ)2{{h}^{2}}={{\left( \sin \theta -\cos \theta \right)}^{2}}+{{\left( \cos \theta +\sin \theta \right)}^{2}}.
⇒h2=sin⁡2θ+cos⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ+sin⁡2θ+2sin⁡θcos⁡θ\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⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 . So, h2=1+1=2{{h}^{2}}=1+1=2 .
⇒h=2\Rightarrow h=\sqrt{2}
Hence, the length of the hypotenuse is 2\sqrt{2} . Hence, option B. is the correct option.

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