A^2+b^2=c^2 so 4(4) +3(3) is 16+9= 25 and the square root of that is 5.
Find <span>tan<span>(<span><span>5π</span>12</span>)</span></span> and sin ((5pi)/12)
Answer: <span>±<span>(2±<span>√3</span>)</span>and±<span><span>√<span>2+<span>√3</span></span></span>2</span></span>
Explanation:
Call tan ((5pi/12) = t.
Use trig identity: <span><span>tan2</span>a=<span><span>2<span>tana</span></span><span>1−<span><span>tan2</span>a</span></span></span></span>
<span><span>tan<span>(<span><span>10π</span>12</span>)</span></span>=<span>tan<span>(<span><span>5π</span>6</span>)</span></span>=−<span>1<span>√3</span></span>=<span><span>2t</span><span>1−<span>t2</span></span></span></span>
<span><span>t2</span>−2<span>√3</span>t−1=0</span>
<span>D=<span>d2</span>=<span>b2</span>−4ac=12+4=16</span>--> <span>d=±4</span>
<span>t=<span>tan<span>(<span><span>5π</span>12</span>)</span></span>=<span><span>2<span>√3</span></span>2</span>±<span>42</span>=2±<span>√3</span></span>
Call <span><span>sin<span>(<span><span>5π</span>12</span>)</span></span>=<span>siny</span></span>
Use trig identity: <span><span>cos2</span>a=1−2<span><span>sin2</span>a</span></span>
<span><span>cos<span>(<span><span>10π</span>12</span>)</span></span>=<span>cos<span>(<span><span>5π</span>6</span>)</span></span>=<span><span>−<span>√3</span></span>2</span>=1−2<span><span>sin2</span>y</span></span>
<span><span><span>sin2</span>y</span>=<span><span>2+<span>√3</span></span>4</span></span>
<span><span>siny</span>=<span>sin<span>(<span><span>5π</span>12</span>)</span></span>=±<span><span><span>√<span>2+<span>√3</span></span></span>2</span></span></span>
Answer:
Step-by-step explanation:
Basically they're both equal
4x-3=x+7
-x. -x
3x-3=7
+3 +3
3x=10
X=10/3
Answer:
q=10
Step-by-step explanation:
-10+10q=9q
-10=-q
q=10
