Trigonometry Trigonometry Math Forum

 May 9th, 2019, 01:04 PM #1 Newbie   Joined: Apr 2019 From: Malawi Posts: 9 Thanks: 0 Math Focus: Trigonometry and calculus Trigonometric Quadrant Suppose that A is a second quadrant angle with sin A = 1/4. Determine the exact value of sin(A/2). The problem has made me crazy. Help please Last edited by Khoxy; May 9th, 2019 at 01:11 PM.
 May 9th, 2019, 01:44 PM #2 Global Moderator   Joined: May 2007 Posts: 6,755 Thanks: 695 $\sin\frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}$ where $\cos A=-\sqrt{\frac{15}{16}}$. The sign for $\cos A$ is negative, since $A$ is in the second quadrant. Ans.=0.984250984251476 Last edited by skipjack; May 9th, 2019 at 01:59 PM.
 May 9th, 2019, 02:02 PM #3 Newbie   Joined: Apr 2019 From: Malawi Posts: 9 Thanks: 0 Math Focus: Trigonometry and calculus Mathman I don't understand how you come up with cos A. Would you clarify?
 May 9th, 2019, 02:10 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,623 Thanks: 2076 cos²(A) ≡ 1 - sin²(A) and cos(A) ≡ 1 - 2sin²(A/2) $\sqrt{\small8 + 2\sqrt{15}}/\small4 = 0.9920$...
 May 10th, 2019, 12:49 PM #5 Global Moderator   Joined: May 2007 Posts: 6,755 Thanks: 695 $\sin{\frac{A}{2}}=0.992029696267167$ Arith. error in previous. $\cos^2A=1-\sin^2A=1-\frac{1}{16}=\frac{15}{16}$. Last edited by skipjack; May 10th, 2019 at 10:35 PM.

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