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Zorluk: OrtaTrigonometric Ratios and Identities

For an acute angle θ\theta, cos⁡(θ)=513\cos(\theta) = \frac{5}{13}. What is the value of 5tan⁡(θ)+13sin⁡(θ)5\tan(\theta) + 13\sin(\theta)?

Cevap: 24

Cevap

24
Using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 with cos⁡(θ)=513\cos(\theta) = \frac{5}{13} gives sin⁡(θ)=1213\sin(\theta) = \frac{12}{13} because θ\theta is an acute angle. The quotient identity gives tan⁡(θ)=sin⁡(θ)cos⁡(θ)=125\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{12}{5}. Substituting these ratios into the expression 5tan⁡(θ)+13sin⁡(θ)5\tan(\theta) + 13\sin(\theta) gives 5(125)+13(1213)=12+12=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 12 + 12 = 24.

Adım Adım Çözüm

1
Find the value of sin⁡(θ)\sin(\theta) using the Pythagorean identity.
sin⁡(θ)=1213\sin(\theta) = \frac{12}{13}
Since sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and θ\theta is an acute angle, the sine value is positive: sin⁡(θ)=1−(513)2=1213\sin(\theta) = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}.
2
Find the value of tan⁡(θ)\tan(\theta) using the quotient identity.
tan⁡(θ)=125\tan(\theta) = \frac{12}{5}
By definition, tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}. Substituting the known values yields tan⁡(θ)=12/135/13=125\tan(\theta) = \frac{12/13}{5/13} = \frac{12}{5}.
3
Substitute the trigonometric ratios into the given expression and simplify.
24
Substituting the values of tan⁡(θ)\tan(\theta) and sin⁡(θ)\sin(\theta) into 5tan⁡(θ)+13sin⁡(θ)5\tan(\theta) + 13\sin(\theta) gives 5(125)+13(1213)=12+12=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 12 + 12 = 24.

Anahtar Kavram

Trigonometric ratios and identities, specifically the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and the definition of tangent as sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)}.

Alternatif Yöntem

Alternatively, draw a right triangle with an acute angle θ\theta. Since cos⁡(θ)=513=adjacenthypotenuse\cos(\theta) = \frac{5}{13} = \frac{\text{adjacent}}{\text{hypotenuse}}, label the adjacent side as 5 and the hypotenuse as 13. By the Pythagorean theorem, the opposite side is 132−52=12\sqrt{13^2 - 5^2} = 12. From this triangle, sin⁡(θ)=oppositehypotenuse=1213\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13} and tan⁡(θ)=oppositeadjacent=125\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5}. Substituting these values into the expression gives 5(125)+13(1213)=245\left(\frac{12}{5}\right) + 13\left(\frac{12}{13}\right) = 24.
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