Answer:
sin(2x) = 12/13
cos(2x) = 5/13
tan(2x) = 12/5
Step-by-step explanation:
cos x = 3/√13, x is in the first quadrant.
Use Pythagorean identity to find sin x.
sin²x + cos²x = 1
sin²x + (3/√13)² = 1
sin²x + 9/13 = 1
sin²x = 4/13
sin x = ±2/√13
Since x is in the first quadrant, sin x = 2/√13.
Use double angle formulas:
sin(2x) = 2 sin x cos x
sin(2x) = 2 (2/√13) (3/√13)
sin(2x) = 12/13
cos(2x) = cos²x − sin²x
cos(2x) = (3/√13)² − (2/√13)²
cos(2x) = 9/13 − 4/13
cos(2x) = 5/13
tan(2x) = sin(2x) / cos(2x)
tan(2x) = (12/13) / (5/13)
tan(2x) = 12/5
The values of the trigonometric functions are required.
[tex]\sin 2x=\dfrac{12}{13}[/tex]
[tex]\cos 2x=\dfrac{5}{13}[/tex]
[tex]\tan 2x=\dfrac{12}{5}[/tex]
The given function is
[tex]\cos x=\dfrac{3}{\sqrt{13}}[/tex]
The trigonometric identities are used here
[tex]\sin x=\sqrt{1-\cos^2 x}\\\Rightarrow \sin x=\sqrt{1-\left(\dfrac{3}{\sqrt{13}}\right)^2}\\\Rightarrow \sin x=\dfrac{2}{\sqrt{13}}[/tex]
[tex]\sin 2x\\ =2\sin x\cos x\\ =2\times \dfrac{2}{\sqrt{13}}\times \dfrac{3}{\sqrt{13}}\\ =\dfrac{12}{13}[/tex]
[tex]\cos 2x\\ =2\cos^2-1\\ =2\left(\dfrac{3}{\sqrt{13}}\right)^2-1\\ =\dfrac{5}{13}[/tex]
[tex]\tan 2x\\ =\dfrac{\sin 2x}{\cos 2x}\\ =\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}\\ =\dfrac{12}{5}[/tex]
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