Trigonometry Problems and Questions with Solutions - Grade 12

Grade 12 trigonometry problems and questions with answers and solutions are presented.

Solve the following questions

1. Prove the identity
   tan²(x) - sin²(x) = tan²(x) sin²(x)
   

   
   

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

   
   

3. Prove the identity
   4 sin(x) cos(x) = sin(4x) / cos(2x)
   

   
   

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

   
   

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

   
   

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

   
   

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

   
   

8. Prove that
   sin(105°) = ( sqrt(6) + sqrt(2) ) / 4
   

   
   

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

   
   

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

    

    .
    
    

Solutions to the Above Problems

1. 
   Use the identity tan(x) = sin(x) / cos(x) in the left hand side of the given identity.
   tan²(x) - sin²(x) = sin²(x) / cos²(x) - sin²(x)
   = [ sin²(x) - cos²(x) sin²(x) ] / cos²(x)
   = sin²(x) [ 1 - cos²(x) ] / cos²(x)
   = sin²(x) sin²(x) / cos²(x)
   = sin²(x) tan²(x) which is equal to the right hand side of the given identity.




2. Use the identities cos(2x) = 2 cos²(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 cos²(x) - 1 ] / [ sin(x) + 2 sin(x) cos(x) ]
   = [ cos(x) + 2 cos²(x) ] / [ sin(x) + 2 sin(x) cos(x) ]
   = cos(x) [1 + 2 cos(x)] / [ sin(x)( 1 + 2 cos(x) ) ]
   = cot(x)




3. 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.




4. 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 to 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




5. 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




6. 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




7. Use the identities sin(2x) = 2 sin(x) cos(x) and cos(2x) = 1 - 2 sin²(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 sin²(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)
   




8. Use the identities sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
   sin(105°) = sin(60° + 45°)
   = sin(60°)cos(45°) + cos(60°) sin(45°)
   = (sqrt(3)/2 )(sqrt(2)/2) + (1/2)(sqrt(2)/2)
   = ( sqrt(6) + sqrt(2) ) / 4




9. If sin(x) = 2/5 then cos(x) = sqrt(1 - (2/5)²) = sqrt(21)/5
   a) Use identity: cos(2x) = 1 - 2 sin²(x) = 17/25
   b) Use identity: cos(4x) = 1 - 2 sin²(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
   



10. 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°) or AC = 20 sin(35°)
    Note that the two internal angles B and C of triangle ABC add up to 90° and therefore the third angle of triangle ABC is a right angle. We can therefore write
    tan(32°) = AB / AC
    Which gives AB = AC tan(32°)
    = 20 sin(35°)tan(32°) = 7.17 ( rounded to 3 significant digits)

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