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secondary 4 | A Maths
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Hi urgently need an answer for this question! Thank you very much
Date Posted: 5 years ago
Views: 284
Thank u :)
sin²2θ(cot²θ - tan²θ)
= (2sinθcosθ)² (cos²θ/sin²θ - sin²θ/cos²θ)
(Remember cotθ = cosθ/sinθ and tanθ = sinθ/cos, so the above are just squared)
= 4sin²θcos²θ ( (cos⁴θ - sin⁴θ)/(sin²θcos²θ) )
= 4(cos⁴θ - sin⁴θ)
= 4(cos²θ + sin²θ)(cos²θ - sin²θ)
(Property used here is a² - b² = (a+b)(a - b) )
= 4(1)(cos2θ)
= 4cos2θ
= (2sinθcosθ)² (cos²θ/sin²θ - sin²θ/cos²θ)
(Remember cotθ = cosθ/sinθ and tanθ = sinθ/cos, so the above are just squared)
= 4sin²θcos²θ ( (cos⁴θ - sin⁴θ)/(sin²θcos²θ) )
= 4(cos⁴θ - sin⁴θ)
= 4(cos²θ + sin²θ)(cos²θ - sin²θ)
(Property used here is a² - b² = (a+b)(a - b) )
= 4(1)(cos2θ)
= 4cos2θ
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done
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clear
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Good evening Candice!!! Here are my workings for this question.
Date Posted:
5 years ago
Thank you
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