Examkin

Soru

Zorluk: KolayFundamental Trigonometric Identities

For an angle θ\theta in the interval π2<θ<π\frac{\pi}{2} < \theta < \pi, the value of cos⁡θ=−35\cos \theta = -\frac{3}{5}. What is the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta?

  1. A
    110\frac{1}{10}
  2. B
    −75-\frac{7}{5}
  3. C
    1
  4. D
    −15-\frac{1}{5}
  5. 15\frac{1}{5}Cevap

Cevap

one-fifth
To evaluate sin⁡θ+cos⁡θ\sin \theta + \cos \theta, we first find the value of sin⁡θ\sin \theta. We substitute cos⁡θ=−35\cos \theta = -\frac{3}{5} into the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, which gives sin⁡2θ+925=1\sin^2 \theta + \frac{9}{25} = 1. Solving for sin⁡2θ\sin^2 \theta yields 1625\frac{16}{25}. Because the angle θ\theta is constrained to the second quadrant (π2<θ<π\frac{\pi}{2} < \theta < \pi), its sine value must be positive, which means sin⁡θ=45\sin \theta = \frac{4}{5}. Adding the values together, we get 45+(−35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.

Adım Adım Çözüm

1
Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the magnitude of the sine function.
sin⁡2θ=1−(−35)2=1−925=1625\sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
The Pythagorean identity relates sine and cosine for any angle.
2
Determine the correct sign of sin⁡θ\sin \theta based on the given quadrant interval.
Since π2<θ<π\frac{\pi}{2} < \theta < \pi, the angle θ\theta lies in Quadrant II, where the sine function is positive. Thus, sin⁡θ=1625=45\sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}.
The trigonometric function values are positive or negative depending on the quadrant on the unit circle.
3
Compute the sum of sin⁡θ\sin \theta and cos⁡θ\cos \theta.
sin⁡θ+cos⁡θ=45+(−35)=15\sin \theta + \cos \theta = \frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}
This gives the final value of the requested expression.

Anahtar Kavram

Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and quadrant rules to calculate trigonometric values.

Alternatif Yöntem

We can sketch a reference right triangle in Quadrant II. Since cos⁡θ=adjacenthypotenuse=−35\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = -\frac{3}{5}, we assign the adjacent side a length of −3-3 along the x-axis and the hypotenuse a length of 55. By the Pythagorean theorem, the opposite vertical side is 52−(−3)2=4\sqrt{5^2 - (-3)^2} = 4. Since the vertical side is in Quadrant II, it is positive. This makes sin⁡θ=oppositehypotenuse=45\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}. Evaluating the sum yields 45+(−35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.
Tahmini Süre:45s
Bu soruyu puanla

Konular

Fundamental Trigonometric Identities

Bu konudan daha fazla soru