Manipulation of trigonometry

Geometry Level pending

Suppose that cos ⁡ 2 x cos ⁡ 2 y = 2 3 % M a t h T y p e ! M T E F ! 2 ! 1 ! + − % f e a a g K a r t 1 e v 2 a q a t C v A U f e B S j u y Z L 2 y d 9 g z L b v y N v 2 C a e r b u L w B L n % h i o v 2 D G i 1 B T f M B a e X a t L x B I 9 g B a e r b d 9 w D Y L w z Y b I t L D h a r q q t u b s r % 4 r N C H b G e a G q i V u 0 J e 9 s q q r p e p C 0 x b b L 8 F 4 r q q r F f p e e a 0 x e 9 L q − J c 9 % v q a q p e p m 0 x b b a 9 p w e 9 Q 8 f s 0 − y q a q p e p a e 9 p g 0 F i r p e p e K k F r 0 x f r − x % f r − x b 9 a d b a q a a e G a c i G a a i a a b e q a a m a a b a a b a a G c b a W a a S a a a e a a c i % G G J b G a a i 4 B a i a a c o h a c a a I Y a G a a m i E a a q a a i G a c o g a c a G G V b G a a i 4 C % a i a a i k d a c a W G 5 b a a a i a b g 2 d a 9 m a a l a a a b a G a a G O m a a q a a i a a i o d a a a % a a a a ! 41 A F ! \frac{{\cos 2x}}{{\cos 2y}} = \frac{2}{3} \% MathType!MTEF!2!1!+- \% feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaci \% GGJbGaai4BaiaacohacaaIYaGaamiEaaqaaiGacogacaGGVbGaai4C \% aiaaikdacaWG5baaaiabg2da9maalaaabaGaaGOmaaqaaiaaiodaaa \% aaaa!41AF!

Find the value of tan ⁡ ( x + y ) × tan ⁡ ( x − y ) % M a t h T y p e ! M T E F ! 2 ! 1 ! + − % f e a a g K a r t 1 e v 2 a q a t C v A U f e B S j u y Z L 2 y d 9 g z L b v y N v 2 C a e r b u L w B L n % h i o v 2 D G i 1 B T f M B a e X a t L x B I 9 g B a e r b d 9 w D Y L w z Y b I t L D h a r q q t u b s r % 4 r N C H b G e a G q i V u 0 J e 9 s q q r p e p C 0 x b b L 8 F 4 r q q r F f p e e a 0 x e 9 L q − J c 9 % v q a q p e p m 0 x b b a 9 p w e 9 Q 8 f s 0 − y q a q p e p a e 9 p g 0 F i r p e p e K k F r 0 x f r − x % f r − x b 9 a d b a q a a e G a c i G a a i a a b e q a a m a a b a a b a a G c b a G a c i i D a i a a c g % g a c a G G U b G a a i i k a i a a d I h a c q G H R a W k c a W G 5 b G a a i y k a i a b g E n a 0 k G a % c s h a c a G G H b G a a i O B a i a a c I c a c a W G 4 b G a e y O e I 0 I a a m y E a i a a c M c a a a % a ! 4627 ! \tan (x + y) \times \tan (x - y) \% MathType!MTEF!2!1!+- \% feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg \% gacaGGUbGaaiikaiaadIhacqGHRaWkcaWG5bGaaiykaiabgEna0kGa \% cshacaGGHbGaaiOBaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcaaa \% a!4627!


The answer is 0.2.

Mas Mus by Mas Mus , 34, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Unstable Chickoy
Jun 19, 2014

Is this solution correct?

c o s 2 x cos ⁡ 2 y = 2 3 = 0.2 0.3 \frac{cos{2x}}{\cos{2y}} = \frac{2}{3} = \frac{0.2}{0.3}

cos ⁡ 2 x = 0.2 \cos{2x} = 0.2

cos ⁡ 2 y = 0.3 \cos{2y} = 0.3

x = 39.23152048 x = 39.23152048

y = 36.27119844 y = 36.27119844

substitute values to

tan ⁡ ( x + y ) × tan ⁡ ( x − y ) = 0.2 \tan(x + y) \times \tan(x - y) = \boxed{0.2}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mas Mus
Apr 12, 2014

Let cos ⁡ 2 x = 2 a \cos2x=2a and cos ⁡ 2 y = 3 a \cos2y=3a

tan ⁡ ( x + y ) × tan ⁡ ( x − y ) = tan ⁡ x + tan ⁡ y 1 − tan ⁡ x tan ⁡ y × tan ⁡ x − tan ⁡ y 1 + tan ⁡ x tan ⁡ y = tan ⁡ 2 x − tan ⁡ 2 y 1 − tan ⁡ 2 x tan ⁡ 2 y = sin ⁡ 2 x cos ⁡ 2 y − sin ⁡ 2 y cos ⁡ 2 x cos ⁡ 2 x cos ⁡ 2 y − sin ⁡ 2 x sin ⁡ 2 y = sin ⁡ 2 x − sin ⁡ 2 y ( sin ⁡ 2 x + cos ⁡ 2 x ) cos ⁡ 2 x − sin ⁡ 2 y ( sin ⁡ 2 x + cos ⁡ 2 x ) = sin ⁡ 2 x − sin ⁡ 2 y cos ⁡ 2 x − sin ⁡ 2 y = 1 2 ( 1 − cos ⁡ 2 x ) − 1 2 ( 1 − cos ⁡ 2 y ) 1 2 ( 1 + cos ⁡ 2 x ) − 1 2 ( 1 − cos ⁡ 2 y ) = − 2 a + 3 a 2 a + 3 a = 1 5 = 0.2 \begin{aligned}\tan(x+y)\times\tan(x-y)&=\frac{\tan x+ \tan y}{1-\tan x \tan y}\times\frac{\tan x-\tan y}{1+\tan x \tan y}\\\\&=\frac{\tan^2x-\tan^2y}{1-\tan^2x \tan^2y}=\frac{\sin^2x \cos^2y-\sin^2y \cos^2x}{\cos^2x \cos^2y-\sin^2x \sin^2y}\\\\&=\frac{\sin^2x - \sin^2y\left(\sin^2x + \cos^2x \right)}{\cos ^2x - \sin^2y\left(\sin^2x + \cos^2x \right)} = \frac{\sin^2x - \sin^2y}{\cos^2x - \sin^2y}\\\\&=\frac{\frac{1}{2}\left(1-\cos2x\right)-\frac{1}{2}\left(1-\cos2y\right)}{\frac{1}{2}\left(1+\cos2x\right)-\frac{1}{2}\left(1-\cos2y\right)}\\\\&=\frac{-2a+3a}{2a+3a}=\frac{1}{5}=\boxed{0.2}\end{aligned}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...