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Zorluk: OrtaFundamental Trigonometric Identities

For an angle θ\theta satisfying π2<θ<π\frac{\pi}{2} < \theta < \pi, if sec⁡θ=−135\sec \theta = -\frac{13}{5}, what is the value of 12(cot⁡θ+csc⁡θ)12(\cot \theta + \csc \theta)?

Cevap: 8

Cevap

8
For an angle θ\theta in Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), cosine is negative and sine is positive. Given sec⁡θ=−135\sec \theta = -\frac{13}{5}, the reciprocal identity gives cos⁡θ=−513\cos \theta = -\frac{5}{13}. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 yields sin⁡θ=1−(−513)2=1213\sin \theta = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \frac{12}{13}. Then cot⁡θ=cos⁡θsin⁡θ=−512\cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{5}{12} and csc⁡θ=1sin⁡θ=1312\csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}. Adding these values gives cot⁡θ+csc⁡θ=812=23\cot \theta + \csc \theta = \frac{8}{12} = \frac{2}{3}. Multiplying by 12 yields the final value of 8.

Adım Adım Çözüm

1
Find cos⁡θ\cos \theta from sec⁡θ\sec \theta
cos⁡θ=−513\cos \theta = -\frac{5}{13}
By definition of the reciprocal trigonometric identity, cos⁡θ=1sec⁡θ\cos \theta = \frac{1}{\sec \theta}.
2
Calculate sin⁡θ\sin \theta using the Pythagorean identity
sin⁡θ=1213\sin \theta = \frac{12}{13}
In Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), sine is positive. Applying sin⁡θ=1−cos⁡2θ\sin \theta = \sqrt{1 - \cos^2 \theta} gives 1−25169=1213\sqrt{1 - \frac{25}{169}} = \frac{12}{13}.
3
Find cot⁡θ\cot \theta and csc⁡θ\csc \theta
\cot \theta = -\frac{5}{12} \text{ and } \csc \theta = \frac{13}{12}
Using quotient identity cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} and reciprocal identity csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}.
4
Substitute into the given expression 12(cot⁡θ+csc⁡θ)12(\cot \theta + \csc \theta) and simplify
8
12(−512+1312)=12(812)=812\left(-\frac{5}{12} + \frac{13}{12}\right) = 12\left(\frac{8}{12}\right) = 8.

Anahtar Kavram

Pythagorean, quotient, and reciprocal identities with quadrant sign analysis
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