If-sin-P-8-17-where-3pi-2-lt-P-lt-2pi-and-cos-Q-24-25-where-pi-lt-Q-lt-3pi-2-what-is-the-value-of-cos-2P-Q-

Question Number 116448 by bobhans last updated on 04/Oct/20

If \sin (P) = - \frac{8}{17}, where \frac{3 π}{2} < P < 2 π

and \cos (Q) = - \frac{24}{25}, where π < Q < \frac{3 π}{2}

what is the value of \cos (2 P + Q) = ?

Answered by bemath last updated on 04/Oct/20

(1) \cos (P) = \frac{15}{17}; \sin (Q) = - \frac{7}{25}

\Rightarrow \cos (2 P + Q) = \cos (2 P) . \cos (Q) - \sin (2 P) . \sin (Q)

= - \frac{24}{25} (1 - 2 (\frac{64}{289})) - {2 (- \frac{8}{17}) (\frac{15}{17})} . (- \frac{7}{25})

= - \frac{24}{25} (\frac{161}{289}) - (\frac{240}{289}) . (\frac{7}{25})

= - \frac{5544}{289 \times 25} = - \frac{5544}{7225}

Answered by mr W last updated on 04/Oct/20

\sin P = - \frac{8}{17}

\Rightarrow \cos P = \frac{\sqrt{17^{2} - 8^{2}}}{17} = \frac{15}{17}

\cos Q = - \frac{24}{25}

\Rightarrow \sin Q = - \frac{\sqrt{25^{2} - 24^{2}}}{25} = - \frac{7}{25}

\cos (2 P + Q) = \cos 2 P \cos Q - \sin 2 P \sin Q

= (2 \cos^{2} P - 1) \cos Q - 2 \sin P \cos P \sin Q

= (2 \times \frac{15^{2}}{17^{2}} - 1) (- \frac{24}{25}) - 2 (- \frac{8}{17} \times \frac{15}{17}) (- \frac{7}{25})

= - \frac{(2 \times 15^{2} - 17^{2}) \times 24 + 2 \times 8 \times 15 \times 7}{17^{2} \times 25}

= - \frac{5544}{7225}

Commented by I want to learn more last updated on 04/Oct/20

Sweet sir

Commented by I want to learn more last updated on 04/Oct/20

Commented by I want to learn more last updated on 04/Oct/20

Sir please tag this your solution . I can only see the image . but i think it is

your solution sir

Commented by mr W last updated on 04/Oct/20

w h a t i s t h e q u e s t i o n ?

Commented by I want to learn more last updated on 04/Oct/20

I only see the image saved sir, no question number, am only thinking it

could be find the radius of the small circle . with the lines you drew on the

diagram

Commented by mr W last updated on 04/Oct/20

s e e Q 116466

Commented by I want to learn more last updated on 04/Oct/20

Am really grateful sir