Expert answers to "For α∈(0∘;90∘) simplify sin2α+tan2α+sin2α⋅cos2α+cos4α My try: sin2α+tan2α+sin2α⋅cos2α+cos4α=sin2α+sin2αcos2α+sin2α⋅cos2α+cos4α =sin2α⋅cos2α+sin2+sin2α⋅cos4α+cos6αcos2α"

Susan Nall

Answered question

2022-01-14

For

α \in (0^{\circ}; 90^{\circ})

simplify

\sin^{2} α + \tan^{2} α + \sin^{2} α \cdot \cos^{2} α + \cos^{4} α

My try:

\sin^{2} α + \tan^{2} α + \sin^{2} α \cdot \cos^{2} α + \cos^{4} α = \sin^{2} α + \frac{\sin^{2} α}{\cos^{2} α} + \sin^{2} α \cdot \cos^{2} α + \cos^{4} α

= \frac{\sin^{2} α \cdot \cos^{2} α + \sin^{2} + \sin^{2} α \cdot \cos^{4} α + \cos^{6} α}{\cos^{2} α}

poleglit3

Beginner2022-01-16Added 32 answers

Express everything in terms of

\cos α = c

(1 - c^{2}) + \frac{(1 - c^{2})}{c^{2}} + (1 - c^{2}) c^{2} + c^{4}

= \frac{c^{2} - c^{4} + 1 - c^{2} + c^{4} - c^{6} + c^{6}}{c^{2}} = \frac{1}{c^{2}}

RizerMix

Expert2022-01-20Added 656 answers

Youre

user_27qwe

Skilled2022-01-23Added 375 answers

Instead of dividing by

\cos^{2} α

, you could do the following,

\sin^{2} α + \tan^{2} α + \sin^{2} α + \sin^{2} α \cos^{2} α + \cos^{4} α

= \sin^{2} α + \tan^{2} α + \sin^{2} α + \cos^{2} α (\sin^{2} α + \cos^{2} α)

= \sin^{2} α + \tan^{2} α + \cos^{2} α

= 1 + \tan^{2} α

= \sec^{2} α

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