
Given $ \cos \theta = \dfrac{{24}}{{25}}$ and $ 270 < \theta < 360,$ how do you find $ \cos \left( {\dfrac{\theta }{2}} \right)?$
Answer
519.6k+ views
Hint: As we know that the above question is related to trigonometric sine cosine is a trigonometric ratio. As we know that the ratio between the adjacent side of the right angled triangle and the hypotenuse of the right angled triangle is called the cosine function. From the question we can see that $ \cos 360$ symbolizes the angle in the fourth quadrant which gives a positive value.
Complete step-by-step answer:
As per the question we have $ \cos \theta = \dfrac{{24}}{{25}}$ and $ 270 < \theta < 360$ . We can solve it by using the double angle formula which is $ \cos 2x = 2{\cos ^2}x - 1$ .
Now we will solve it for $ \cos x$ which gives the half angle formula i.e. by dividing both sides by $ 2$ $ \cos x = \pm \sqrt {\dfrac{1}{2}\left( {\cos 2x + 1} \right)} $ .
Therefore we can write the above formula as $ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2}\left( {\cos \theta + 1} \right)} $ . Now we will put the values and solve it as
$ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2}\left( {\dfrac{{24}}{{25}} + 1} \right)} $ .
This gives us
$ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2} \times \dfrac{{49}}{{25}}}\\
\Rightarrow \pm \sqrt {\dfrac{{49}}{{50}}} $ .
In this question we are taking $ \theta $ as a positive angle in the fourth quadrant but its half angle between $ 135$ and $ 180$ lies in the second quadrant. So this gives us the negative cosine. So the value will be$ \cos \dfrac{\theta }{2} = - \sqrt {\dfrac{{49}}{{50}}} $ .
We can simplify it by multiplying both the numerator and denominator with $ 2$ i.e. $ - \sqrt {\dfrac{{2(49)}}{{100}}} $ . It gives us $ - \dfrac{7}{{10}}\sqrt 2 $ .
Hence the required answer is $ \cos \dfrac{\theta }{2} = - \dfrac{7}{{10}}\sqrt 2 $ .
So, the correct answer is “$ \cos \dfrac{\theta }{2} = - \dfrac{7}{{10}}\sqrt 2 $ ”.
Note: Before solving this type of question we should have the proper knowledge of trigonometric ratios and their formulas. Also we should keep in mind the quadrant as the value of the cosine depends on this. The value of cosine lies positive in the first and the fourth quadrant while it’s value is negative in the second and the third quadrant.
Complete step-by-step answer:
As per the question we have $ \cos \theta = \dfrac{{24}}{{25}}$ and $ 270 < \theta < 360$ . We can solve it by using the double angle formula which is $ \cos 2x = 2{\cos ^2}x - 1$ .
Now we will solve it for $ \cos x$ which gives the half angle formula i.e. by dividing both sides by $ 2$ $ \cos x = \pm \sqrt {\dfrac{1}{2}\left( {\cos 2x + 1} \right)} $ .
Therefore we can write the above formula as $ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2}\left( {\cos \theta + 1} \right)} $ . Now we will put the values and solve it as
$ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2}\left( {\dfrac{{24}}{{25}} + 1} \right)} $ .
This gives us
$ \cos \dfrac{\theta }{2} = \pm \sqrt {\dfrac{1}{2} \times \dfrac{{49}}{{25}}}\\
\Rightarrow \pm \sqrt {\dfrac{{49}}{{50}}} $ .
In this question we are taking $ \theta $ as a positive angle in the fourth quadrant but its half angle between $ 135$ and $ 180$ lies in the second quadrant. So this gives us the negative cosine. So the value will be$ \cos \dfrac{\theta }{2} = - \sqrt {\dfrac{{49}}{{50}}} $ .
We can simplify it by multiplying both the numerator and denominator with $ 2$ i.e. $ - \sqrt {\dfrac{{2(49)}}{{100}}} $ . It gives us $ - \dfrac{7}{{10}}\sqrt 2 $ .
Hence the required answer is $ \cos \dfrac{\theta }{2} = - \dfrac{7}{{10}}\sqrt 2 $ .
So, the correct answer is “$ \cos \dfrac{\theta }{2} = - \dfrac{7}{{10}}\sqrt 2 $ ”.
Note: Before solving this type of question we should have the proper knowledge of trigonometric ratios and their formulas. Also we should keep in mind the quadrant as the value of the cosine depends on this. The value of cosine lies positive in the first and the fourth quadrant while it’s value is negative in the second and the third quadrant.
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