Question

If theta is an acute angle and 3sin theta =4cos theta find the value of 4sin ^2-3cos ^2 + 2

Answer 1

Answer:

sin theta/cos theta=4/3

tan theta =4/3= p/b

sin theta=p/h

=4/5

cos theta=b/h

=3/5

4*16/25-3*9/25+2=64/25-27/25+2

=37/25+2

=3.48

is the answer right??

Answer 2

Given is the relation between sine and cosine for an angle, Find $4sin^2\theta-3cos^2\theta+2$

Explanation:

• In a right angled triangle having perpendicular 'P', base length 'B', and hypotenuse 'H' w.r.t an angle $\theta$ we have, $->tan\theta=\frac{P}{B},\ sin\theta=\frac{P}{H} \ \ \ \ \ \ \ \ \ ->H=\sqrt{P^2+B^2}$    ----(a)
• Here we have, $3sin\theta=4cos\theta\ \ \ \ ->tan\theta=\frac{3}{4}$  
• Now from (a) we get,
• $P=3,\ \ \ B=4\ \ \ \\\ ->H=\sqrt{3^2+4^2}=5 \\->sin\theta=\frac{3}{5}$     ----(b)
• Now we have to evaluate,
• $4sin^2\theta-3cos^2\theta+2\\->4sin^2\theta-3(1-sin^2\theta)+2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (->sin^2\theta+cos^2\theta=1)\\->7sin^2\theta-1\\->7(\frac{3}{5}) ^2-1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (->from\ (b))\\->\frac{63}{25} -1\\\\->\frac{38}{25}= 1.52$  
• Hence, the value of the required expression is $\frac{38}{25}\ (=1.52)$.