IB Math SL Trigonometric Identities Questions

Question 1 :

Show that 

tan²θ - sin²θ = tan²θsin²θ

Solution :

tan²θ - sin²θ = tan²θ - sin²θ(cos²θ/cos²θ)

= tan²θ - (sin²θ/cos²θ)cos²θ

= tan²θ - tan²θcos²θ

= tan²θ(1 - cos²θ)

= tan²θsin²θ

Question 2 :

Show that 

secθ - cosθ = tanθsinθ

Solution :

secθ - cosθ = 1/cosθ - cosθ

= (1 - cos²θ)/cosθ

= sin²θ/cosθ

= (sinθ/cosθ)sinθ

= tanθsinθ

Question 3 :

Show that

sin2x + cos2x - 1 = 2sinx(cosx - sinx)

Answer :

sin2x + cos2x - 1 :

= 2sinxcosx + cos²x - sin²x - sin²x - cos²x

= 2sinxcosx - sin²x - sin²x

= 2sinxcosx - 2sin²x

= 2sinx(cosx - sinx)

Question 4 :

Show that 

cos⁴θ - sin⁴θ = cos2θ

Solution :

cos⁴θ - sin⁴θ = (cos²θ)² - (sin²θ)²

Using the identity a² - b² = (a + b)(a - b),  

= (cos²θ + sin²θ)(cos²θ - sin²θ)

= (1)(cos2θ)

= cos2θ

Question 5 :

Show that 

sin⁴θ + cos⁴θ = 1 - 2sin²θcos²θ

Solution :

sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)²

Using the identity a² + b² = (a + b)² - 2ab,  

= (sin²θ + cos²θ)² - 2sin²θcos²θ

= 1² - 2sin²θcos²θ

= 1 - 2sin²θcos²θ

Question 6 :

Show that the equation 2cos²x + 5sinx = 4 may be written in the form

2sin²x - 5sinx + 2 = 0

Answer :

2cos²x + 5sinx = 4

2(1 - sin²x) + 5sinx = 4

2 - 2sin²x + 5sinx = 4

-2sin²x + 5sinx + 2 = 4

Multiply both sides by -1.

-1(-2sin²x + 5sinx + 2) = -1(4)

2sin²x - 5sinx - 2 = -4

Add 4 to both sides.

2sin²x - 5sinx + 2 = 0

Question 7 :

Show that 

Answer :

Question 8 :

Show that 

Answer :

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