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Math  /  Trigonometry

QuestionA. π2\frac{\pi}{2} B. 3π2\frac{3 \pi}{2} C. π\pi D.undefined Q.27) The value of 4sin⁡−1(519)+4cos⁡−1(519)=4 \sin ^{-1}\left(\frac{\sqrt{5}}{19}\right)+4 \cos ^{-1}\left(\frac{\sqrt{5}}{19}\right)= A. π\pi B. 12π\frac{1}{2} \pi C. 14π\frac{1}{4} \pi D. 2π2 \pi

Studdy Solution

STEP 1

1. The problem involves inverse trigonometric functions.
2. The identity sin⁡−1(x)+cos⁡−1(x)=π2 \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} can be used for any x x in the domain of the inverse sine and cosine functions.

STEP 2

1. Apply the identity for inverse trigonometric functions.
2. Simplify the expression using the identity.
3. Calculate the final value.

STEP 3

Recall the identity for inverse trigonometric functions:
sin⁡−1(x)+cos⁡−1(x)=π2 \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}
This identity holds true for any x x such that −1≤x≤1 -1 \leq x \leq 1 .

STEP 4

Apply the identity to the given expression:
4sin⁡−1(519)+4cos⁡−1(519) 4 \sin^{-1}\left(\frac{\sqrt{5}}{19}\right) + 4 \cos^{-1}\left(\frac{\sqrt{5}}{19}\right)
Using the identity:
sin⁡−1(519)+cos⁡−1(519)=π2 \sin^{-1}\left(\frac{\sqrt{5}}{19}\right) + \cos^{-1}\left(\frac{\sqrt{5}}{19}\right) = \frac{\pi}{2}
Multiply both sides of the identity by 4:
4(sin⁡−1(519)+cos⁡−1(519))=4×π2 4 \left( \sin^{-1}\left(\frac{\sqrt{5}}{19}\right) + \cos^{-1}\left(\frac{\sqrt{5}}{19}\right) \right) = 4 \times \frac{\pi}{2}

STEP 5

Simplify the expression:
4×π2=2π 4 \times \frac{\pi}{2} = 2\pi
The value of the expression is:
2π \boxed{2\pi}

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