A civil engineer is designing a ramp for a building. While checking structural calculations, she uses trigonometric expressions to model stress ratios and load distribution at a given angle θ.
A civil engineer is designing a ramp for a building. While checking the structural calculations, she writes the following trigonometric expression that models the stress ratio at angle θ:
(sin θ + cos θ)² + (sin θ − cos θ)²
She also needs to verify a second identity used in load-distribution calculations:
(tan²θ − sin²θ) = tan²θ · sin²θ
(i) Evaluate the engineer's stress-ratio expression: (sin θ + cos θ)² + (sin θ − cos θ)² [1 mark]
(ii) Using the identity sin²θ + cos²θ = 1, prove the load-distribution identity:
(tan²θ − sin²θ) = tan²θ · sin²θ [3 marks]
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(sin θ + cos θ)² + (sin θ − cos θ)²
⟹ [sin²θ + 2 sin θ cos θ + cos²θ] + [sin²θ − 2 sin θ cos θ + cos²θ]
⟹ [sin²θ + cos²θ + 2 sin θ cos θ] + [sin²θ + cos²θ − 2 sin θ cos θ]
Using the identity sin²θ + cos²θ = 1:
⟹ [1 + 2 sin θ cos θ] + [1 − 2 sin θ cos θ]
⟹ 1 + 1
∴ (sin θ + cos θ)² + (sin θ − cos θ)² = 2
(ii) To Prove: tan²θ − sin²θ = tan²θ · sin²θ
Taking the Left Hand Side (LHS):
LHS = tan²θ − sin²θ
⟹ sin²θ/cos²θ − sin²θ [∵ tan θ = sin θ/cos θ]
⟹ sin²θ · (1/cos²θ − 1) [taking sin²θ common]
⟹ sin²θ · ((1 − cos²θ)/cos²θ)
Using the identity sin²θ + cos²θ = 1, we have 1 − cos²θ = sin²θ:
⟹ sin²θ · (sin²θ/cos²θ)
⟹ sin²θ · tan²θ
⟹ tan²θ · sin²θ
= RHS. Hence proved.