Question

If tan A =4/3 then find the values of the following:
i) sin 2A
ii) cos 2A
iii) tan 2A​

Answer 1

Answer:

1) sin2A = 2tanx/1+tan^2x

[2×4/3] / [(1+(4/3)^2]

(8/3)/[25/9]

sin 2A = 24/25

2) tan a = 4/3.

tan 2a = 2 tan a/(1 - tan^2 a)

= [2*4/3]/[1 - (4/3)^2]

= (8/3)/[ 1- 16/9]

= (8/3)/[-7/9]

= (8/3)*(-9/7)

tan 2a= (-)24/7

3) cos2A= 1-tan^2A/1+tan^2A

= [2×4/3]\[1-(4/3)^2]

=(-7/9)/(25/9)

= (-)7/25

Answer 2

Values are as follows - sin 2A is 24/25, cos 2A is -7/25 and tan 2A is -24/7.

Given - Tan A

Find - sin 2A, cos 2A and tan 2A

Solution - As we know, tan = perpendicular/base

So, the perpendicular and base will be 4 and 3. Finding Hypotenuse using Pythagoras theorem.

${Hypotenuse}^{2} = {4}^{2} + {3}^{2}$

${Hypotenuse}^{2} = 16 + 9$

${Hypotenuse}^{2} = 25$

$Hypotenuse = \sqrt{25}$

$Hypotenuse = 5$

sin = perpendicular/hypotenuse.

$\sin( A ) = \frac{4}{5}$

cos = base/hypotenuse

$\cos( A) = \frac{3}{5}$

tan A = $\frac{4}{3}$

Now, sin 2A = 2 sinA cos A

cos 2A = 2 cos² A - 1

tan 2A = 2 tan A/(1- tan² A)

Finding sin 2A

(i) sin 2A = $2 \times \frac{4}{5} \times \frac{3}{5}$

(i) sin 2A = $\frac{24}{25}$

(ii) cos 2A = $2 \times( { \frac{3}{5}) }^{2} - 1$

(ii) cos 2A = $2 \times \frac{9}{25} - 1$

(ii) cos 2A = $\frac{18}{25} - 1$

(ii) cos 2A = $\frac{18 - 25}{25}$

(ii) cos 2A = $\frac{ - 7}{25}$

(iii) tan 2A = $\frac{2 \times \frac{4}{3} }{(1 - {( \frac{4}{3}) }^{2} }$

tan 2A = $\frac{ \frac{8}{3} }{(1 - \frac{16}{9}) }$

tan 2A = $\frac{ \frac{8}{3} }{( \frac{9 - 16}{9}) }$

tan 2A = $\frac{ \frac{8}{3} }{( \frac{ - 7}{9}) }$

tan 2A = $- \frac{8 \times 9}{7 \times 3}$

tan 2A = $- \frac{72}{21}$

tan 2A = -24/7

Hence, sin 2A is 24/25, cos 2A is -7/25 and tan 2A is -24/7.

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