Trigonometry! #31

Geometry Level 4

Find the value of cos ⁡ ( 2 cos ⁡ − 1 x + sin ⁡ − 1 x ) \cos(2\cos^{-1}x + \sin^{-1}x) at x = 1 5 x=\frac {1}{5} , where 0 ≤ cos ⁡ − 1 x ≤ π 0≤\cos^{-1}x≤\pi and − π 2 ≤ sin ⁡ − 1 x ≤ π 2 \frac {-\pi}{2} ≤ \sin^{-1}x ≤ \frac {\pi}{2} .

If your answer is in the form of − a b c -\frac {a\sqrt{b}}{c} , where b b is a square free integer, enter the value of a + b + c a+b+c .

This problem is part of the set Trigonometry .


The answer is 13.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Chew-Seong Cheong
Jan 18, 2015

Let α = cos ⁡ − 1 1 5 \space\alpha = \cos ^{-1} {\frac{1}{5}}\space and β = = sin ⁡ − 1 1 5 \space \beta = = \sin ^{-1} {\frac{1}{5}} . Then, we have:

cos ⁡ ( 2 cos ⁡ − 1 1 5 + sin ⁡ − 1 1 5 ) = cos ⁡ ( 2 α + β ) \cos {(2\cos ^{-1} {\frac{1}{5}}+ \sin^{-1} {\frac{1}{5}})} = \cos {(2\alpha+\beta)}

= cos ⁡ 2 α cos ⁡ β − sin ⁡ 2 α sin ⁡ β \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \cos {2\alpha}\cos{\beta} - \sin {2\alpha}\sin{\beta}

= ( cos ⁡ 2 α − 1 ) cos ⁡ β − 2 sin ⁡ α cos ⁡ α sin ⁡ β \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (\cos^2 {\alpha}-1)\cos{\beta} - 2\sin {\alpha}\cos{\alpha}\sin{\beta}

= ( 1 25 − 1 ) cos ⁡ β − 2 ( 1 25 ) sin ⁡ α \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (\frac{1}{25}-1)\cos{\beta} - 2(\frac{1}{25})\sin {\alpha}

= ( 1 25 − 1 ) 1 − 1 25 − 2 25 1 − 1 25 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = (\frac{1}{25}-1)\sqrt{1-\frac{1}{25}} - \frac{2}{25}\sqrt{1-\frac{1}{25}}

= ( − 23 25 − 2 25 ) 24 25 = − 2 6 5 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = \left( -\frac{23}{25} - \frac{2}{25} \right) \sqrt{\frac{24}{25}} = - \frac {2\sqrt{6}}{5}

⇒ a + b + c = 2 + 6 + 5 = 13 \Rightarrow a + b + c = 2+6+5 = \boxed{13}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

c o s ( π 2 + c o s − 1 x ) = − s i n ( c o s − 1 x ) = − 1 − x 2 cos( \dfrac{\pi}{2} + cos^{-1}x) = -sin(cos^{-1}x) = - \sqrt{1 - x^2}

= − 2 6 5 = \dfrac{-2\sqrt{6}}{5}

U Z - 6 years, 4 months ago

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