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Category: Trigonometry

Question-199480

Question Number 199480 by Abdullahrussell last updated on 04/Nov/23 Commented by mr W last updated on 04/Nov/23 $$=\frac{\mathrm{sin}\:\theta+\mathrm{sin}\:\mathrm{50}°+\mathrm{1}+\mathrm{sin}\:\mathrm{50}°}{\mathrm{cos}\:\theta+\mathrm{cos}\:\mathrm{50}°+\mathrm{0}−\mathrm{cos}\:\mathrm{50}°} \\ $$$$=\frac{\mathrm{sin}\:\theta+\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\mathrm{50}°}{\mathrm{cos}\:\theta} \\ $$ Terms of Service…

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Question Number 199424 by cortano12 last updated on 03/Nov/23 $$\:\:\:\boldsymbol{{x}} \\ $$ Answered by Frix last updated on 04/Nov/23 $${f}\left({x}\right)=\frac{\mathrm{cos}\:{x}}{\mathrm{3}}\left(\mathrm{6sin}^{\mathrm{3}} \:{x}\:−\mathrm{4sin}^{\mathrm{2}} \:{x}\:+\mathrm{1}\right) \\ $$$${f}'\left({x}\right)=−\mathrm{sin}\:{x}\:\left(\mathrm{1}−\mathrm{2sin}\:{x}\right)\left(\mathrm{3}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)…