

Prove the indentity
tan^{2}(x)  sin^{2}(x) = tan^{2}(x) sin^{2}(x)

Prove the indentity
(1 + cos(x) + cos(2x)) / (sin(x) + sin(2x)) = cot(x)

Prove the indentity
4 sin(x) cos(x) = sin(4x) / cos(2x)

Solve the trigonometric equation given by
sin(x) + sin(x/2) = 0 for 0 ≤ x ≤ 2 pi

Solve the trigonometric equation given by
(2sin(x)  1)(tan(x)  1) = 0 for 0 ≤ x ≤ 2 pi

Solve the trigonometric equation given by
cos(2x) cos(x)  sin(2x) sin(x) = 0 for 0 ≤ x ≤ 2 pi

Solve the trigonometric equation given by
( sin(2x)  cos(x) ) / ( cos(2x) + sin(x)  1 ) = 0 for 0 ≤ x ≤ 2 pi

Prove that
sin(105^{o}) = ( sqrt(6) + sqrt(2) ) / 4

If sin(x) = 2/5 and x is an accute angle, find the exact values of
a) cos(2x)
b) cos(4x)
c) sin(2x)
d) sin(4x)

Find the length of side AB in the figure below. Round your answer to 3 significant digits.
.
Solutions to the Above Problems

Use the identity tan(x) = sin(x) / cos(x) in the left hand side of the given identity.
tan^{2}(x)  sin^{2}(x) = sin^{2}(x) / cos^{2}(x)  sin^{2}(x)
= [ sin^{2}(x)  cos^{2}(x) sin^{2}(x) ] / cos^{2}(x)
= sin^{2}(x) [ 1  cos^{2}(x) ] / cos^{2}(x)
= sin^{2}(x) sin^{2}(x) / cos^{2}(x)
= sin^{2}(x) tan^{2}(x) which is equal to the right hand side of the given identity.

Use the identities cos(2x) = 2 cos^{2}(x)  1 and sin(2x) = 2 sin(x) cos(x) in the left hand side of the given identity.
[ 1 + cos(x) + cos(2x) ] / [ sin(x) + sin(2x) ]
= [ 1 + cos(x) + 2 cos^{2}(x)  1 ] / [ sin(x) + 2 sin(x) cos(x) ]
= [ cos(x) + 2 cos^{2}(x) ] / [ sin(x) + 2 sin(x) cos(x) ]
= cos(x) [1 + 2 cos(x)] / [ sin(x)( 1 + 2 cos(x) ) ]
= cot(x)

Use the identity sin(2x) = 2 sin(x) cos(x) to write sin(4x) = 2 sin(2x) cos(2x) in the right hand side of the given identity.
sin(4x) / cos(2x)
= 2 sin(2x) cos(2x) / cos(2x)
= 2 sin(2x)
= 2 [ 2 sin(x) cos(x)]
= 4 sin(x) cos(x) which is equal to the right hand side of the given identity.

Use the identity sin(2x) = 2 sin(x) cos(x) to write sin(x) as sin(2 * x/2) = 2 sin(x / 2) cos(x / 2) in the right hand side of the given equation.
2 sin(x / 2) cos(x / 2) + sin(x / 2) = 0
sin(x/2) [ 2 cos(x/2) + 1 ] = 0 factor
which gives
sin(x/2) = 0 or 2 cos(x/2) + 1 = 0
sin(x / 2) = 0 leads to x / 2 = 0 or x / 2 = Pi which leads ro x = 0 or x = 2pi
2 cos(x/2) + 1 = 0 leads to cos(x/2) = 1/2 which leads to x/2 = 2pi/3 and x/2 = 4pi/3 (the second solution leads to x greater than 2pi)
solutions: x = 0, x = 4pi/3 and x = 2pi

The given equation is already factored
(2sin(x)  1)(tan(x)  1) = 0
which means
2sin(x)  1 = 0 or tan(x)  1 = 0
sin(x) = 1/2 or tan(x) = 1 equivalent equations to the above
solutions: x = pi/6, 5pi/6, x = pi/4 and x = 5pi/4

Note that cos(2x + x) = cos(2x) cos(x)  sin(2x) sin(x) using the formula for cos(A + B). Hence
cos(2x) cos(x)  sin(2x) sin(x) = 0 is equivalent to
cos(3x) = 0
Solve for 3x to obtain: 3x = pi/2, 3x = 3Pi/2, 3x = 5pi/2, 3x = 7pi/2, 3x = 9pi/2 and 11pi/2
solutions: x = pi/6, pi/2, 5pi/6, 7pi/6, 3pi/2 and 11pi/6

Use the identities sin(2x) = 2 sin(x) cos(x) and cos(2x) = 1  2 sin^{2}(x) to rewrite the given equation as follows
given equation
( sin(2x)  cos(x) ) / ( cos(2x) + sin(x)  1 ) = 0
( 2 sin(x) cos(x)  cos(x) ) / ( 1  2 sin^{2}(x) + sin(x)  1) = 0
cos(x)( 2 sin(x)  1 ) / [  sin(x)( 2 sin(x)  1 ) ] = 0
Divide numerator and denominator by 2 sin(x)  1 to simplify; assuming that 2 sin(x)  1 is not equal to zero.
 cos(x) / sin(x) = 0
cot(x) = 0
solutions: x = pi/2 and x = 3pi/2
We now need to verify that both solutions found make neither the denominator nor 2 sin(x)  1 equal to zero. (do this as an exercise)

Use the identities sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
sin(105^{o}) = sin(60^{o} + 45^{o})
= sin(60^{o})cos(45^{o}) + cos(60^{o}) sin(45^{o})
= (sqrt(3)/2 )(sqrt(2)/2) + (1/2)(sqrt(2)/2)
= ( sqrt(6) + sqrt(2) ) / 4

If sin(x) = 2/5 then cos(x) = sqrt(1  (2/5)^{2}) = sqrt(21)/5
a) Use identity: cos(2x) = 1  2 sin^{2}(x) = 17/25
b) Use identinty: cos(4x) = 1  2 sin^{2}(2 x)
= 1  2 [ 2sin(x) cos(x) ]^{2}
= 457 / 625
c) sin(2x) = 2 sin(x) cos(x) = 4 sqrt(21)/25
d) sin(4x) = sin(2(2x)) = 2 cos(2x) sin(2x)
= 2 (17/25)(4 sqrt(21)/25) = 136 sqrt(21) / 625

Note that triangle DAC is isosceles and therefore if we draw the perpendicular from D to AC, it will cut AC into two halves and bisect angle D. Hence
(1/2) AC = 10 sin(35^{o}) or AC = 20 sin(35^{o})
Note that the two internal angles B and C of triangle ABC add up to 90^{o} and therefore the third angle of triangle ABC is a right angle.
We can therefore write
tan(32^{o}) = AB / AC
Which gives AB = AC tan(32^{o})
= 20 sin(35^{o})tan(32^{o}) = 7.17 ( rounded to 3 significant digits)

