Thread: Trigonometric Relations/Identities help

Please help me solve these:

1a) If tan(A) = 3/4, show that sin(A) = 3/4cos(A)

1b) Therefore, evaluate:
[4sin(A) - cos(A)] / [8sin(A) + 5cos(A)]

2) Solve without a calculator, show working
a) sin(60) x tan(45)
b) 9tan^2(30)
c) 6cos(60) x sin(30)

3) If sin^2(x) = 7/8, find cos^2(x)

4) If cos^2(x) = 0.36, fins sin^2(x)

5) Find tan x if cos x = 8/17 and sin x = 15/17

The last one is as an optional challenge, please help me if you can.

6) ACD is a right angled triangle, where B is a point on AC, angle CAD = 30 deg, angle CBD = 60 deg and AB = 10.

a) By considering angle BCD, prove CD = BC
b) By considering angle ACD, prove CD = [10 + BC] / [SQRT 3]
c) Use these two results to find BC

Thank you in advance!

Hello,

Originally Posted by BG5965

Please help me solve these:

1a) If tan(A) = 3/4, show that sin(A) = 3/4cos(A)

Use the definition of tan :

1b) Therefore, evaluate:
[4sin(A) - cos(A)] / [8sin(A) + 5cos(A)]

Substitute

For example,

2) Solve without a calculator, show working
a) sin(60) x tan(45)
b) 9tan^2(30)
c) 6cos(60) x sin(30)

Use this unit circle : http://dcr.csusb.edu/LearningCenter/...UnitCircle.gif

3) If sin^2(x) = 7/8, find cos^2(x)

4) If cos^2(x) = 0.36, fins sin^2(x)

Use identity

5) Find tan x if cos x = 8/17 and sin x = 15/17

Use the definition of tan I gave you above

The last one is as an optional challenge, please help me if you can.

6) ACD is a right angled triangle, where B is a point on AC, angle CAD = 30 deg, angle CBD = 60 deg and AB = 10.

a) By considering angle BCD, prove CD = BC
b) By considering angle ACD, prove CD = [10 + BC] / [SQRT 3]
c) Use these two results to find BC

Draw a sketch and use definitions of cosine, sine and tangent :

cos=adjacent side/hypotenuse
sin=opposit side/hypotenuse
tan=opposit side/adjacent side

If you don't know it, search on wikipedia

Hi,

2.)

sin60*tan45

=(sqrt3)/2 * 1

=(sqrt3)/2

b.)

=9tan^2(30)

=9(1/sqrt3)^2
=9(1/3)
=3

c.)

=6Cos60sin30
=6(1/2)(1/2)
=3/2