Question

If tan theta = 3/4 then find the value of 2 sin theta + 3 cos theta.

Answer 1

Given
$\tan( \alpha ) = \frac{3}{4}$
$2 \sin( \alpha ) + 3 \cos( \alpha ) = x$
Divide the equation by
$\cos( \alpha )$
$2 \tan( \alpha ) + 3 = x \sec( \alpha )$
Square both side
${(2 \tan( \alpha ) + 3)}^{2} = {(x \sec( \alpha ) )}^{2}$
${(2 \times \frac{3}{4} + 3 )}^{2} = {x}^{2} {(1 + { \tan( \alpha ) }^{2} ) )}$
${( \frac{9}{2} )}^{2} = {x}^{2} {(1 + \frac{9}{16} )}$
${x}^{2} = \frac{81}{16} \times \frac{4}{25}$
${x}^{2} = \frac{81×16}{25×4}$
$x = \sqrt{ \frac{81×16}{25×4} } = + - \frac{18}{5}$

Answer 2

The value of $\bf2 \sin( \theta) + 3 \cos( \theta)$ is 18/5.

Given:

• $\tan( \theta) = \frac{3}{4}$

To find:

• Value of $2 \sin( \theta) + 3 \cos( \theta)$

Solution:

Formula to be used:

• $tan \theta = \frac{\text{side \: opposite \: to \: angle}}{\text{side \: adjacent\: to \: angle}}$
• $sin \theta = \frac{\text{side \: opposite \: to \: angle}}{hypotenuse}$
• $cos\theta = \frac{\text{side \: adjacent \: to \: angle}}{hypotenuse}$

Step 1:

Find hypotenuse of right triangle.

As,

Ratio of base to perpendicular is given so find hypotenuse.

i.e.

$\tan( \theta) =\frac{AB}{AC}= \frac{3k}{4k}$

Apply Pythagoras theorem

Hypotenuse² = Base² + Perpendicular² $CB^2= 9 {k}^{2} + 16 {k}^{2}$

Hypotenuse (CB)= 5k

Step 2:

Find sin theta form formula.

$\sin( \theta) = \frac{AB}{CB} =\frac{3k}{5k}$

or

$\bf \sin( \theta) = \frac{3}{5}$

Step 3:

Find cos theta from formula.

$\cos( \theta) = \frac{AC}{CB}= \frac{4k}{5k}$

or

$\bf \cos( \theta) = \frac{4}{5}$

Step 4:

Put the values in expression

$= 2 \times \frac{3}{5} + 3 \times \frac{4}{5}$

or

$= \frac{6}{5} + \frac{12}{5}$

or

$2 \sin( \theta) + 3 \cos( \theta) = \frac{18}{5}$

Thus,

$\bf 2 \sin( \theta) + 3 \cos( \theta) = \frac{18}{5}$

Learn more:

1. If sin x =3/5 and cos y =12/13, evaluate:

i)sec2 x

ii)tan x + tan y

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2) if 21 cosec theta =29 find the value of cossquare theta-sinsquare theta/1-2sinsquare theta

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