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Question : 112 of 180
Marks:
+1,
-0
Solution:
To find the value of
sin‌−1[cos(39‌)], we start by simplifying the argument of the cosine function.
First, notice that
39‌ can be simplified using the periodicity of the cosine function which is
2Ï€. That is,
cos(θ)=cos(θ+2πk) for any integer (k).
We observe that:
39‌=‌=7π+‌This can be further simplified in multiples of
(2Ï€) :
So,
cos(39‌)=cos(π+‌)Note that
cos(π+x)=−cos(x), hence
cos(π+‌)=−cos(‌)Since
cos(π−x)=−cos(x), and
π−‌=‌, and
−cos(‌)=cos(‌), we find that:
cos(39‌)=cos(‌)Now, we evaluate
sin‌−1[cos(‌)].
By properties of trigonometric inverse functions and identities, particularly the fact that:
sin‌(‌−x)=cos(x)we have:
Thus, the correct answer is Option B
‌.
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