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Zorluk: OrtaBasic Trigonometric Ratios, Special Angles, and Identities

Given that sin⁡θ=513\sin \theta = \frac{5}{13}, where θ\theta is an acute angle, evaluate the value of 13cos⁡θ−12tan⁡θ13 \cos \theta - 12 \tan \theta. What is the numerical value?

Cevap: 7

Cevap

The numerical value of the expression 13cos⁡θ−12tan⁡θ13 \cos \theta - 12 \tan \theta is 7.
For an acute angle θ\theta with sin⁡θ=513\sin \theta = \frac{5}{13}, the corresponding right triangle has an opposite side of 5, a hypotenuse of 13, and an adjacent side of 132−52=12\sqrt{13^2 - 5^2} = 12. Therefore, cos⁡θ=1213\cos \theta = \frac{12}{13} and tan⁡θ=512\tan \theta = \frac{5}{12}. Evaluating 13cos⁡θ−12tan⁡θ13 \cos \theta - 12 \tan \theta yields 13(1213)−12(512)=12−5=713\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7.

Adım Adım Çözüm

1
Determine cos⁡θ\cos \theta using the right triangle ratio or Pythagorean identity.
cos⁡θ=1213\cos \theta = \frac{12}{13}
Since sin⁡θ=oppositehypotenuse=513\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}, the adjacent side is 132−52=12\sqrt{13^2 - 5^2} = 12. Because θ\theta is acute, cos⁡θ\cos \theta is positive.
2
Determine tan⁡θ\tan \theta using the ratio of opposite to adjacent sides.
tan⁡tan⁡θ=512\tan \tan \theta = \frac{5}{12}
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}$.
3
Substitute the evaluated ratios into 13cos⁡θ−12tan⁡θ13 \cos \theta - 12 \tan \theta and simplify.
13\left(\frac{12}{13}\right) - 12\left(\frac{5}{12}\right) = 12 - 5 = 7
Multiplying clears the denominators, leaving 12−5=712 - 5 = 7.

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Basic Trigonometric Ratios and Pythagorean Triples
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Basic Trigonometric Ratios, Special Angles, and Identities

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