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

If sin (P) = −(8/(17)) ,where ((3π)/2)<P<2π and cos (Q)=−((24)/(25)), where π<Q<((3π)/2) what is the value of cos (2P+Q) =?

$$\mathrm{If}\:\mathrm{sin}\:\left(\mathrm{P}\right)\:=\:−\frac{\mathrm{8}}{\mathrm{17}}\:,\mathrm{where}\:\frac{\mathrm{3}\pi}{\mathrm{2}}<\mathrm{P}<\mathrm{2}\pi \\ $$ $$\mathrm{and}\:\mathrm{cos}\:\left(\mathrm{Q}\right)=−\frac{\mathrm{24}}{\mathrm{25}},\:\mathrm{where}\:\pi<\mathrm{Q}<\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$ $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cos}\:\left(\mathrm{2P}+\mathrm{Q}\right)\:=? \\ $$

Answered by bemath last updated on 04/Oct/20

(1) cos (P)= ((15)/(17)) ; sin (Q)=−(7/(25)) ⇒cos (2P+Q) = cos (2P).cos (Q)−sin (2P).sin (Q) =−((24)/(25))(1−2(((64)/(289))))−{2(−(8/(17)))(((15)/(17)))}.(−(7/(25))) = −((24)/(25))(((161)/(289)))−(((240)/(289))).((7/(25))) = −((5544)/(289×25)) = −((5544)/(7225))

$$\left(\mathrm{1}\right)\:\mathrm{cos}\:\left(\mathrm{P}\right)=\:\frac{\mathrm{15}}{\mathrm{17}}\:;\:\mathrm{sin}\:\left(\mathrm{Q}\right)=−\frac{\mathrm{7}}{\mathrm{25}} \\ $$ $$\Rightarrow\mathrm{cos}\:\left(\mathrm{2P}+\mathrm{Q}\right)\:=\:\mathrm{cos}\:\left(\mathrm{2P}\right).\mathrm{cos}\:\left(\mathrm{Q}\right)−\mathrm{sin}\:\left(\mathrm{2P}\right).\mathrm{sin}\:\left(\mathrm{Q}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{24}}{\mathrm{25}}\left(\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{64}}{\mathrm{289}}\right)\right)−\left\{\mathrm{2}\left(−\frac{\mathrm{8}}{\mathrm{17}}\right)\left(\frac{\mathrm{15}}{\mathrm{17}}\right)\right\}.\left(−\frac{\mathrm{7}}{\mathrm{25}}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{24}}{\mathrm{25}}\left(\frac{\mathrm{161}}{\mathrm{289}}\right)−\left(\frac{\mathrm{240}}{\mathrm{289}}\right).\left(\frac{\mathrm{7}}{\mathrm{25}}\right) \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{5544}}{\mathrm{289}×\mathrm{25}}\:=\:−\frac{\mathrm{5544}}{\mathrm{7225}} \\ $$

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

sin P=−(8/(17)) ⇒cos P=((√(17^2 −8^2 ))/(17))=((15)/(17)) cos Q=−((24)/(25)) ⇒sin Q=−((√(25^2 −24^2 ))/(25))=−(7/(25)) cos (2P+Q)=cos 2P cos Q−sin 2P sin Q =(2 cos^2 P−1) cos Q−2 sin P cos P sin Q =(2×((15^2 )/(17^2 ))−1)(−((24)/(25)))−2(−(8/(17))×((15)/(17)))(−(7/(25))) =−(((2×15^2 −17^2 )×24+2×8×15×7)/(17^2 ×25)) =−((5544)/(7225))

$$\mathrm{sin}\:{P}=−\frac{\mathrm{8}}{\mathrm{17}} \\ $$ $$\Rightarrow\mathrm{cos}\:{P}=\frac{\sqrt{\mathrm{17}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} }}{\mathrm{17}}=\frac{\mathrm{15}}{\mathrm{17}} \\ $$ $$\mathrm{cos}\:{Q}=−\frac{\mathrm{24}}{\mathrm{25}} \\ $$ $$\Rightarrow\mathrm{sin}\:{Q}=−\frac{\sqrt{\mathrm{25}^{\mathrm{2}} −\mathrm{24}^{\mathrm{2}} }}{\mathrm{25}}=−\frac{\mathrm{7}}{\mathrm{25}} \\ $$ $$\mathrm{cos}\:\left(\mathrm{2}{P}+{Q}\right)=\mathrm{cos}\:\mathrm{2}{P}\:\mathrm{cos}\:{Q}−\mathrm{sin}\:\mathrm{2}{P}\:\mathrm{sin}\:{Q} \\ $$ $$=\left(\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{P}−\mathrm{1}\right)\:\mathrm{cos}\:{Q}−\mathrm{2}\:\mathrm{sin}\:{P}\:\mathrm{cos}\:{P}\:\mathrm{sin}\:{Q} \\ $$ $$=\left(\mathrm{2}×\frac{\mathrm{15}^{\mathrm{2}} }{\mathrm{17}^{\mathrm{2}} }−\mathrm{1}\right)\left(−\frac{\mathrm{24}}{\mathrm{25}}\right)−\mathrm{2}\left(−\frac{\mathrm{8}}{\mathrm{17}}×\frac{\mathrm{15}}{\mathrm{17}}\right)\left(−\frac{\mathrm{7}}{\mathrm{25}}\right) \\ $$ $$=−\frac{\left(\mathrm{2}×\mathrm{15}^{\mathrm{2}} −\mathrm{17}^{\mathrm{2}} \right)×\mathrm{24}+\mathrm{2}×\mathrm{8}×\mathrm{15}×\mathrm{7}}{\mathrm{17}^{\mathrm{2}} ×\mathrm{25}} \\ $$ $$=−\frac{\mathrm{5544}}{\mathrm{7225}} \\ $$

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

Sweet sir

$$\mathrm{Sweet}\:\mathrm{sir} \\ $$

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

Commented byI 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

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{tag}\:\mathrm{this}\:\mathrm{your}\:\mathrm{solution}.\:\mathrm{I}\:\mathrm{can}\:\mathrm{only}\:\mathrm{see}\:\mathrm{the}\:\mathrm{image}.\:\mathrm{but}\:\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is} \\ $$ $$\mathrm{your}\:\mathrm{solution}\:\mathrm{sir} \\ $$

Commented bymr W last updated on 04/Oct/20

what is the question?

$${what}\:{is}\:{the}\:{question}? \\ $$

Commented byI 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

$$\mathrm{I}\:\mathrm{only}\:\mathrm{see}\:\mathrm{the}\:\mathrm{image}\:\mathrm{saved}\:\mathrm{sir},\:\mathrm{no}\:\mathrm{question}\:\mathrm{number},\:\mathrm{am}\:\mathrm{only}\:\mathrm{thinking}\:\mathrm{it} \\ $$ $$\mathrm{could}\:\mathrm{be}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{small}\:\mathrm{circle}.\:\mathrm{with}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{you}\:\mathrm{drew}\:\mathrm{on}\:\mathrm{the} \\ $$ $$\mathrm{diagram} \\ $$

Commented bymr W last updated on 04/Oct/20

see Q116466

$${see}\:{Q}\mathrm{116466} \\ $$

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

Am really grateful sir

$$\mathrm{Am}\:\mathrm{really}\:\mathrm{grateful}\:\mathrm{sir} \\ $$

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