Question

tan(arcsin3/5 + arccos5/7 ) =?

Answer 1

find with calculator the arcsin by shift+sin (5/3) and arccos by shift+cos (...)

Answer 2

Then add them and then compute the tan of this sum result

Answer 3

we cant use calculator..

Answer 4

I think it is an artificial "smartass" problem. And i openly admit I cannot solve it w.o. ca

Answer 5

i have my answer but im not yet sure if its correct ill just compare it to yours..

Answer 6

Ah now I realized - You have to use addition formula for tan\[\tan(\alpha + \beta) = ....\]

Answer 7

yeah.. hehe

Answer 8

\[= \frac{ \tan \alpha + \tan \beta }{ 1 - \tan \alpha \tan \beta }\]

Answer 9

tan(a + b) = sin(a + b)/cos(a + b)
= (sin a cos b + cos a sin b)/(cos a cos b - sin a sin b)

Divide both numerator and denominator by cos a cos b

tan (a + b) = (tan a + tan b)/(1 - tan a tan b)

Answer 10

Also U will have to fnd their complementary functions: by Pythagoras equality I gave you before \[\sin^2 \alpha + \cos^2 \alpha = 1\]

Answer 11

Actualy what ali usedis even faster
\[\tan(a + b) = \sin(a + b)/\cos(a + b) = (\sin a \cos b + \cos a \sin b)/(\cos a \cos b - \sin a \sin b) \]

Answer 12

\[ (\sin a \cos b + \cos a \sin b)/(\cos a \cos b - \sin a \sin b) \]

Answer 13

thanks guys.. what do you think is the final answer?

Answer 14

1 Find the complementary to sin and to cos
2 substitute in the LAST formula that I have printed from Ali-s advice (just above ur last question

Answer 15

This one \[(\sin a \cos b + \cos a \sin b)/(\cos a \cos b - \sin a \sin b) \]

Answer 16

\[\sin x = \sqrt{1 - \cos^2 x}\]

Answer 17

\[\tan(\arcsin3/5 + \arccos5/7 ) = ?\]

Answer 18

the answer should be in fraction form when you already find the a and b of tan and it has square roots..

Answer 19

Pls figure out how to use the sqrt I wrote for You and then substitute into this\[(\sin a \cos b + \cos a \sin b)/(\cos a \cos b - \sin a \sin b)\]

Answer 20

ok sure tnx

Answer 21

LEARN FROM BOOK WHAT IS INVERSE TRIGONOMETRIC FUNCTION - it is the angle actually