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____ [38]
4 years ago
5

Give the exact value of the expression without using a calculator.

Mathematics
1 answer:
Art [367]4 years ago
8 0

Answer:

\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}.

Step-by-step explanation:

To find the exact value of the expression \sin(\tan^{-1}(\frac{8}{15} ))

First, we need to simplify the expression \sin(\tan^{-1}(x)).

Draw a triangle in the plane with vertices (1,x), (1,0), and the origin. Then \tan^{-1}(x) is the angle between the positive x-axis and the ray beginning at the origin and passing through (1,x).

Therefore,

\sin(\tan^{-1}(x))=\frac{x}{\sqrt{1+x^{2} } }

\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}

\frac{x\sqrt{1+x^2}}{\sqrt{1+x^2}\sqrt{1+x^2}}

\sqrt{1+x^2}\sqrt{1+x^2}=1+x^2

\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}

Now, use the identity \sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}

\sin(\tan^{-1}(\frac{8}{15} ))=\frac{\left(\frac{8}{15}\right)\sqrt{1+\left(\frac{8}{15}\right)^2}}{1+\left(\frac{8}{15}\right)^2}\\\\\frac{\frac{8}{15}\sqrt{\left(\frac{8}{15}\right)^2+1}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{64}{225}}\\\\\frac{136}{225\cdot \frac{289}{225}}\\\\\frac{136}{289}=\frac{8}{17}\\\\\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}

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