Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Suppose that sin alpha=(3)/(sqrt13) and 0^(@) < alpha < 90^(@) Find the exact values of cos ((alpha)/(2)) and tan ((alpha)/(2)). cos ((alpha)/(2))= tan ((alpha)/(2))=

Suppose that sin⁡α=313 \sin \alpha=\frac{3}{\sqrt{13}} and 0^{\circ}<\alpha<90^{\circ} \newlineFind the exact values of cos⁡α2 \cos \frac{\alpha}{2} and tan⁡α2 \tan \frac{\alpha}{2} .\newlinecos⁡α2= \cos \frac{\alpha}{2}= \newlinetan⁡α2= \tan \frac{\alpha}{2}=

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Evaluate functionsright chevron icon

Full solution

Q. Suppose that sin⁡α=313 \sin \alpha=\frac{3}{\sqrt{13}} and 0∘<α<90∘ 0^{\circ}<\alpha<90^{\circ} \newlineFind the exact values of cos⁡α2 \cos \frac{\alpha}{2} and tan⁡α2 \tan \frac{\alpha}{2} .\newlinecos⁡α2= \cos \frac{\alpha}{2}= \newlinetan⁡α2= \tan \frac{\alpha}{2}=
  1. Given Information: We are given that sin⁡(α)=313\sin(\alpha) = \frac{3}{\sqrt{13}} and α\alpha is in the first quadrant (0^\circ < \alpha < 90^\circ). To find cos⁡(α/2)\cos(\alpha/2) and tan⁡(α/2)\tan(\alpha/2), we can use the half-angle formulas for sine and cosine. The half-angle formula for cosine is:\newlinecos⁡(α/2)=±(1+cos⁡(α)2)\cos(\alpha/2) = \pm\sqrt{\left(\frac{1 + \cos(\alpha)}{2}\right)}\newlineSince α\alpha is in the first quadrant, cos⁡(α)\cos(\alpha) will be positive, and since we are looking for cos⁡(α/2)\cos(\alpha/2) in the first or second quadrant (because α/2\alpha/2 will be less than 90∘90^\circ), we will choose the positive square root.\newlineFirst, we need to find cos⁡(α)\cos(\alpha) using the Pythagorean identity:\newlinecos⁡2(α)+sin⁡2(α)=1\cos^2(\alpha) + \sin^2(\alpha) = 1\newlinecos⁡2(α)=1−sin⁡2(α)\cos^2(\alpha) = 1 - \sin^2(\alpha)\newlinecos⁡2(α)=1−(313)2\cos^2(\alpha) = 1 - \left(\frac{3}{\sqrt{13}}\right)^2\newlinecos⁡2(α)=1−913\cos^2(\alpha) = 1 - \frac{9}{13}\newlinecos⁡2(α)=1313−913\cos^2(\alpha) = \frac{13}{13} - \frac{9}{13}\newlinecos⁡2(α)=413\cos^2(\alpha) = \frac{4}{13}\newlineα\alpha11\newlineα\alpha22
  2. Find cos⁡(α/2)\cos(\alpha/2): Now we can find cos⁡(α/2)\cos(\alpha/2) using the half-angle formula:\newlinecos⁡(α/2)=±(1+cos⁡(α))/2\cos(\alpha/2) = \pm\sqrt{(1 + \cos(\alpha))/2}\newlinecos⁡(α/2)=(1+2/13)/2\cos(\alpha/2) = \sqrt{(1 + 2/\sqrt{13})/2}\newlinecos⁡(α/2)=(13/13+2/13)/2\cos(\alpha/2) = \sqrt{(\sqrt{13}/\sqrt{13} + 2/\sqrt{13})/2}\newlinecos⁡(α/2)=(13+2)/213\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)/2\sqrt{13}}\newlinecos⁡(α/2)=(13+2)/(213)\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)/(2\sqrt{13})}\newlinecos⁡(α/2)=(13+2)13/(21313)\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)\sqrt{13}/(2\sqrt{13}\sqrt{13})}\newlinecos⁡(α/2)=(13+2)13/(26)\cos(\alpha/2) = \sqrt{(\sqrt{13} + 2)\sqrt{13}/(26)}\newlinecos⁡(α/2)=(13+213)/26\cos(\alpha/2) = \sqrt{(13 + 2\sqrt{13})/26}\newlinecos⁡(α/2)=(13+213)/26\cos(\alpha/2) = \sqrt{(13 + 2\sqrt{13})/26}
  3. Find tan⁡(α/2)\tan(\alpha/2): Next, we will find tan⁡(α/2)\tan(\alpha/2) using the half-angle formula for tangent, which is:\newlinetan⁡(α/2)=±(1−cos⁡(α))/(1+cos⁡(α))\tan(\alpha/2) = \pm\sqrt{(1 - \cos(\alpha))/(1 + \cos(\alpha))}\newlineSince α\alpha is in the first quadrant, tan⁡(α)\tan(\alpha) will be positive, and we will choose the positive square root for tan⁡(α/2)\tan(\alpha/2).\newlinetan⁡(α/2)=(1−cos⁡(α))/(1+cos⁡(α))\tan(\alpha/2) = \sqrt{(1 - \cos(\alpha))/(1 + \cos(\alpha))}\newlinetan⁡(α/2)=(1−2/13)/(1+2/13)\tan(\alpha/2) = \sqrt{(1 - 2/\sqrt{13})/(1 + 2/\sqrt{13})}\newlinetan⁡(α/2)=(13/13−2/13)/(13/13+2/13)\tan(\alpha/2) = \sqrt{(\sqrt{13}/\sqrt{13} - 2/\sqrt{13})/(\sqrt{13}/\sqrt{13} + 2/\sqrt{13})}\newlinetan⁡(α/2)=(13−2)/(13+2)\tan(\alpha/2) = \sqrt{(\sqrt{13} - 2)/(\sqrt{13} + 2)}\newlinetan⁡(α/2)\tan(\alpha/2)00\newlinetan⁡(α/2)\tan(\alpha/2)11\newlinetan⁡(α/2)\tan(\alpha/2)11

More problems from Evaluate functions

Question
What is the domain of this function?\newline(9,–2)(6,–10)(–3,9)(5,2)(9,–2)(6,–10)(–3,9)(5,2)\newlineChoices:\newline(A){–3,5,6,9}(B){–10,–2,2,9}(C){2,5,6,9}(D){–6,–3,5,9}(A)\{–3,5,6,9\}\newline(B)\{–10,–2,2,9\}\newline(C)\{2,5,6,9\}\newline(D)\{–6,–3,5,9\}
Get tutor helpright-arrow

Posted 7 months ago

Question
Look at this set of ordered pairs:\newline(2,14)(2, 14)\newline(10,12)(10, 12)\newline(3,−19)(3, -19)\newlineIs this relation a function?\newlineChoices:\newline[[yes]\text{[[yes]}[no]]\text{[no]]}
Get tutor helpright-arrow

Posted 7 months ago

Question
Use the following function rule to find f(11) f(11) .\newlinef(x)=−10−7x f(x) = -10 - 7x \newlinef(11)= f(11) = _____
Get tutor helpright-arrow

Posted 7 months ago

Question
Find the slope of the line y=−3x−25 y = -3x - \frac{2}{5} .\newlineSimplify your answer and write it as a proper fraction, improper fraction, or integer.\newline_____\_\_\_\_\_
Get tutor helpright-arrow

Posted 7 months ago

Question
A line has a slope of 33 and passes through the point (−2,−10)(-2,-10). Write its equation in slope-intercept form.\newlineWrite your answer using integers, proper fractions, and improper fractions in simplest form.
Get tutor helpright-arrow

Posted 7 months ago

Question
What is the range of this function?\newliney = |x|\newlineChoices:\newlineall real numbers\text{all real numbers}\newline{y∣y≥0}\{y \mid y \geq 0\}\newline{y∣y≤0}\{y \mid y \leq 0\}\newline{y∣y>0}\{y \mid y > 0\}
Get tutor helpright-arrow

Posted 9 months ago

Question
Find g(x) g(x) , where g(x) g(x) is the translation 1 1 unit down of f(x)=∣x∣ f(x) = |x| .\newlineWrite your answer in the form a∣x−h∣+k a|x - h| + k , where a a , h h , and k k are integers.\newlineg(x)= g(x) = ______\newline
Get tutor helpright-arrow

Posted 9 months ago

Question
What is the domain of this function?\newline(8,–1)(7,–11)(–4,10)(4,3)(8,–1)(7,–11)(–4,10)(4,3)\newlineChoices: \newline(A){–4,4,7,8}(A)\{–4,4,7,8\}\newline(B){–11,–1,3,10}(B)\{–11,–1,3,10\}\newline(C){3,4,7,8}(C)\{3,4,7,8\}\newline(D){–7,–4,4,8}(D)\{–7,–4,4,8\}
Get tutor helpright-arrow

Posted 9 months ago

Question
What is the domain of this function?\newline(10, –3)(2, –9)(–5, 8)(3, 1)(10,\,–3)(2,\,–9)(–5,\,8)(3,\,1)\newlineChoices:\newline[{–5,2,3,10}][{–9, –3,1,8}][{1,2,3,10}][{–10, –5,2,3}][\{–5,2,3,10\}][\{–9,\,–3,1,8\}][\{1,2,3,10\}][\{–10,\,–5,2,3\}]
Get tutor helpright-arrow

Posted 7 months ago

Question
What is the domain of this function?\newline(1,–4)(9,–12)(–6,11)(7,5)(1,–4)(9,–12)(–6,11)(7,5)\newlineChoices:(A){–6,1,7,9}\newline\\(A)\{–6,1,7,9\}\newline(B){–12,–4,5,11}(B)\{–12,–4,5,11\}\newline(C{5,1,7,9}(C\{5,1,7,9\}\newline(D){–9,–6,1,7}(D)\{–9,–6,1,7\}
Get tutor helpright-arrow

Posted 7 months ago

View all (97)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant