Trigonometry

Geometry Level 2

If sec ⁡ A = 17 8 \sec A = \dfrac{17}8 , find 3 − 4 sin ⁡ 2 A 4 cos ⁡ 2 A − 3 \dfrac{3-4\sin^2 A}{4\cos^2 A - 3} .

33 611 \frac{33}{611} 1 1 30 4 \frac{30}4 0 0

Vishwathiga Jayasankar by vishwathiga jayasankar , 20, India

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1 solution

3 − 4 sin ⁡ 2 A 4 cos ⁡ 2 A − 3 = 3 − 4 ( 1 − cos ⁡ 2 A ) 4 cos ⁡ 2 A − 3 = 4 cos ⁡ 2 A − 1 4 cos ⁡ 2 A − 3 Divide up and down by cos ⁡ 2 A = 4 − sec ⁡ 2 A 4 − 3 sec ⁡ 2 A = 4 − ( 17 8 ) 2 4 − 3 ( 17 8 ) 2 = 256 − 289 256 − 867 = − 33 − 611 = 33 611 \begin{aligned} \frac {3-4\sin^2A}{4\cos^2 A-3} & = \frac {3-4(1-\cos^2A)}{4\cos^2 A-3} \\ & = \frac {4\cos^2A -1}{4\cos^2 A-3} & \small \color{#3D99F6}{\text{Divide up and down by }\cos^2 A} \\ & = \frac {4 -\sec^2 A}{4-3\sec^2 A} \\ & = \frac {4 -\left(\frac {17}8\right)^2}{4-3\left(\frac {17}8\right)^2} \\ & = \frac {256-289}{256 -867} \\ & = \frac {-33}{-611} \\ & = \boxed{\dfrac {33}{611}} \end{aligned}

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sir you are the king of solutions

Sathvik Acharya - 4 years, 3 months ago

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