# Given that sin theta cos theta =7/50 and 0^0 < theta < 90^0. Find the value of cos theta.?

## Help pls...

Jun 22, 2018

$\frac{7}{5 \cdot \sqrt{2}} , \frac{1}{5 \cdot \sqrt{2}}$

#### Explanation:

Solving the equation
$\sin \left(\theta\right) \cdot \cos \left(\theta\right) = \frac{7}{50}$
in the given interval we get

${x}_{1} = - 2 \arctan \left(7 - 5 \sqrt{2}\right)$

${x}_{2} = 2 \arctan \left(\frac{1}{7} \left(5 \sqrt{2} - 1\right)\right)$
so we gat

$\cos \left({x}_{1}\right) = \frac{7}{5 \sqrt{2}}$

$\cos \left({x}_{2}\right) = \frac{1}{5 \sqrt{2}}$

Jun 22, 2018

${8}^{\circ} 13 , {81}^{\circ} 87$

#### Explanation:

$\sin t . \cos t = \frac{7}{50}$
Use trig identity: sin 2t = 2sin t.cos t
In this case:
$\sin t . \cos t = \frac{\sin 2 t}{2} = \frac{7}{50}$
$\sin 2 t = \frac{15}{50} = \frac{7}{25}$
Calculator and unit circle give 2 solutions for 2t:
$2 t = {16}^{\circ} 26$, and $2 t = 180 - 16.26 = {163}^{\circ} 74$
a. 2t = 16.26 + k360
$t = {8}^{\circ} 13 + k {180}^{\circ}$
b. 2t = 163.74 + k360
$t = {81}^{\circ} 87 + k {180}^{\circ}$
${8}^{\circ} 13 , {81}^{\circ} 87$