Question

trigonometry question

Answer 1

given \[ (\sin \theta + \cos \theta)^{2}=3/2 \]
when \[ 0\le \theta \le \pi/4 \]
find the value of
\[\cos^{-1} (\tan 3\theta)\]

Answer 2

\[\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 3/2\]\[\sin^2 \theta + \cos^2 \theta = 1 \]\[2\sin \theta \cos \theta = \sin (2\theta)\]\[\sin (2 \theta) + 1 = 3/2\]\[\sin(2\theta) = 1/2\]

Answer 3

solve for theta then substitute it back to \[\cos^{-1} (\tan 3\theta)\]

Answer 4

I got the same answer as moneybird but went about it a different way

Answer 5

\[2\sin \theta \cos \theta =\frac{1}{2}\]\[\sin \theta cos \theta=\frac{1}{4}\]

So \[\tan3\theta=\frac{\sin3 \theta}{\cos3 \theta}\] \[\tan3\theta=\frac{3\sin \theta-4sin^3\theta}{4cos^3\theta-3\cos \theta}\]\[\tan3\theta=\frac{3\sin \theta-4sin\theta(1-cos^2\theta)}{4cos\theta(1-sin^2\theta)-3\cos \theta}\]\[\tan3\theta=\frac{3\sin \theta-4sin\theta+4sin \theta cos \theta cos\theta}{4cos\theta-4sin \theta cos \theta sin \theta)-3\cos \theta}\]\[\tan3\theta=\frac{3\sin \theta-4sin\theta+cos\theta}{4cos\theta-sin \theta-3\cos \theta}\]\[\tan3\theta=\frac{-\sin \theta+cos\theta}{cos\theta-sin \theta}\]\[\tan3\theta=1\]

So \[\cos^{-1} (1) = 0\]

Answer 6

thx a lot

Answer 7

no worries, good luck

Answer 8

\[\text{ArcCos }\left[\text{ Tan }\left[3 \text{ ArcCos }\left[\frac{\sqrt{2+\sqrt{3}}}{2}\right]\right]\right]=0 \]

Answer 9

GRATE WORK