Maurice Maldonado

2022-04-29

If $\mathrm{sin}2x=\frac{5}{13}$ and $0}^{\circ}<x<{45}^{\circ$ , find $\mathrm{sin}x$ and $\mathrm{cos}x$ .

louran20z47

Beginner2022-04-30Added 14 answers

Suppose we had a right triangle with an angle $2x$ , and $\mathrm{sin}2x=\frac{5}{13}$ . Further suppose that the hypotenuse of the triangle was 13. We can deduce that the side oppoites $2x$ must be 5. Applying the Pythagorean theorem to find the other side we have

$13}^{2}={5}^{2}+{\left(\text{adjacent side}\right)}^{2$

$169-25={\left(\text{adjacent side}\right)}^{2}$

$144={\left(\text{adjacent side}\right)}^{2}$

implying that the side opposite angle 2x is 12. This allows us to state that

$\mathrm{cos}2x=\frac{12}{13}$

Which is easier to work with because

$\mathrm{cos}2x={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=2{\mathrm{cos}}^{2}x-1$

Substituting we have

$\frac{12}{13}=2{\mathrm{cos}}^{2}x-1$

$\mathrm{cos}}^{2}x=\frac{25}{26$

$\mathrm{cos}x=\frac{5}{\sqrt{26}}=\frac{5\sqrt{26}}{26}$

Applying the Pythagorean identity we have

${\mathrm{cos}}^{2}x+{\mathrm{sin}}^{2}x=1$

$\frac{25}{26}+{\mathrm{sin}}^{2}x=1$

$\mathrm{sin}}^{2}x=\frac{1}{26$

$\mathrm{sin}x=\frac{1}{\sqrt{26}}=\frac{\sqrt{26}}{26}$

implying that the side opposite angle 2x is 12. This allows us to state that

Which is easier to work with because

Substituting we have

Applying the Pythagorean identity we have

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