Question

given that tanθ=2/3 and θ is in the 3rd quadrant.without evaluating the angle θ,find the exact values of

a)cos(-θ)
b)sin(-θ)
c)sin(θ-8π)

Answer 1

My Mind has got rusted.. :)

Answer 2

really
thanks for the help

Answer 3

Use :

\[1 + \tan^2(\theta) = \sec^2(\theta)\]

Here find sec(theta) first...

Answer 4

But theta here is in third quadrant, so sec(theta) will be negative..

Answer 5

\[\sec(\theta) = -\sqrt{\frac{5}{3}}\]

Answer 6

Sorry that is not 4, that is 5 there..

Answer 7

Just take the reciprocal of sec(theta) to find cos(theta) ie.

\[\cos(\theta) = \frac{1}{\sec(\theta)}\]

Now using :
\[\sin(\theta) = \cos(\theta) \cdot \tan(\theta)\]

Find sin(theta)..

Answer 8

the ans for a is cos(-θ)=-3/Square root 13
the ans for b is sin(-θ)=2/Square root 13
the ans for c is sin(θ-8π)=-2/Square root 13

i got the answer but i not sure how to do

Answer 9

@waterineyes \[\tan ^{2} (\theta)=(\frac{2}{3})^{2}=\frac{4}{9}\]

Answer 10

I said my mind has got rusty.. :) @kropot72

Answer 11

\[1 + (\frac{2}{3})^2 = \sec^2(\theta) \implies \sec(\theta) = 1 + \frac{4}{9} \implies \sec^2(\theta) = \frac{13}{9}\]

Answer 12

Now find sed(theta) here..

\[\sec(\theta) = \sqrt{\frac{13}{9}} \implies \frac{\sqrt{13}}{3}\]