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Question

(i) If cos θ=45 and θ is acute, find tan 2θ

(ii) If sin θ=45 and 0 < θ < π2, find the value of sin 4 θ

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Solution

(i) Given:
cos θ=45
Thus,
tanθ=sinθcosθ=1-cos2θcosθ=1-45245 =1-162545=34 Because θ lies in the 1st quadrant, tanθ will be positive.Now, tan2θ=2×341-342 tan2θ=2tanθ1-tan2θ =321-916=247

(ii) Given:
sin θ=45 and 0 < θ < π2

Thus,

cosθ=1-sin2θ=1-452=35 0<θ<π2We know, sin2θ=2sinθcosθsin2θ=2×45×35=2425

Again,cos2θ=1-sin22θ=1-24252=±725Because θ lies in the 1st quadrant and 2θ lies in the 2nd quadrant, cos2θ will be negative.cos2θ=-725Now,We know, sin4θ=2sin2θcos2θsin4θ=2×2425×-725=-336625

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