How do you find the exact value of $\mathrm{sin}105$ degrees?

Francisco Proctor
2022-07-22
Answered

How do you find the exact value of $\mathrm{sin}105$ degrees?

You can still ask an expert for help

Clarissa Adkins

Answered 2022-07-23
Author has **16** answers

Use $\mathrm{sin}105=\mathrm{sin}(60+45)=\mathrm{sin}60\mathrm{cos}45+\mathrm{cos}60\mathrm{sin}45$

$=(\frac{\sqrt{3}}{2})(\frac{1}{\sqrt{2}})+(\frac{1}{2})(\frac{1}{\sqrt{2}})=\frac{\sqrt{2}}{4}((\sqrt{3}+1)=0.9656$ nearly.

Explanation:

$\mathrm{sin}45,\mathrm{cos}45$ and $\mathrm{sin}60$ are irrational.

$=(\frac{\sqrt{3}}{2})(\frac{1}{\sqrt{2}})+(\frac{1}{2})(\frac{1}{\sqrt{2}})=\frac{\sqrt{2}}{4}((\sqrt{3}+1)=0.9656$ nearly.

Explanation:

$\mathrm{sin}45,\mathrm{cos}45$ and $\mathrm{sin}60$ are irrational.

Talon Mcbride

Answered 2022-07-24
Author has **2** answers

We know $\mathrm{sin}(A+B)=\mathrm{sin}A\mathrm{cos}B+\mathrm{cos}A\mathrm{sin}B$

Hence $\mathrm{sin}{105}^{\circ}$

$=\mathrm{sin}({60}^{\circ}+{45}^{\circ})$

$=\mathrm{sin}{60}^{\circ}\mathrm{cos}{45}^{\circ}+\mathrm{cos}{60}^{\circ}\mathrm{sin}{45}^{\circ}$

$=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}$

$=\frac{\sqrt{3}+1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{6}+\sqrt{2}}{4}$

Hence $\mathrm{sin}{105}^{\circ}$

$=\mathrm{sin}({60}^{\circ}+{45}^{\circ})$

$=\mathrm{sin}{60}^{\circ}\mathrm{cos}{45}^{\circ}+\mathrm{cos}{60}^{\circ}\mathrm{sin}{45}^{\circ}$

$=\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{1}{\sqrt{2}}$

$=\frac{\sqrt{3}+1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{\sqrt{6}+\sqrt{2}}{4}$

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