Question

If sin(α) = 24/25 where 0 < α < π/2 and cos(β) = 21/29 where 3π/2 < β < 2π.
sin(α) = 24/25
sin(β) = -7/25
cos(α)= 20/29
cos(β)= 21/29
how do you do tan(α − β)?
i already know the formula but i can't get the right answer

Answer 1

well you used the wrong formula

Answer 2

the formula is \(\sin^2\theta+\cos^2\theta=1\)

Answer 3

i think i used the right one

Answer 4

\[\sin\alpha=\dfrac{24}{25}\]\[\cos\alpha=\dfrac7{25}\]
\[\cos\beta=\dfrac{21}{29}\]\[\sin\beta=\dfrac{20}{29}\]

Answer 5

how can \(\cos\beta\) be positive if \(\dfrac{3\pi}2<\beta<2\pi\)?

Answer 6

oh. sorry can you help me solve it.

Answer 7

Oh sorry let me do that again

Answer 8

okay

Answer 9

\[\sin\alpha=\frac{24}{25}\]\[cos\alpha=\sqrt{1-\sin^2\alpha}=\frac7{25}\]
\[\cos\beta=\frac{21}{29}\]\[\sin\beta=-\sqrt{1-\cos^2\alpha}=-\frac{20}{29}\]

Answer 10

formula: \(\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\)

Answer 11

\[\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{24}7\]
Can you work out \(\tan\beta\) and plug that into the formula? :)

Answer 12

no i'm really having a hard time right now figuring it out

Answer 13

\(\tan\beta\) is \(\dfrac{\sin\beta}{\cos\beta}\) now can you work out the value of it?

Answer 14

-20/21 right?

Answer 15

yep :)