Simplifying $dfrac{cot(34^circ)sin(44^circ)}{sin(22^circ)sin(56^circ)} $
$begingroup$
$$dfrac{cot(34^circ)sin(44^circ)}{sin(22^circ)sin(56^circ)} $$
Simplify given expression.
We know that $cot alpha = dfrac{sin alpha }{cos alpha }$ and $sin(44^circ) = 2sin(22^circ)cos(22^circ)$
$$dfrac{cot(34^circ) 2sin(22^circ)cos(22^circ)}{sin(22^circ)sin(56^circ)}$$
Cancelling similar terms out
$$dfrac{cot(34^circ)2cos (22^circ)}{sin(56^circ)} = dfrac{dfrac{cos(34^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)} $$
Using the identity $cos (90^circ-alpha ) = sin (alpha)$
$$dfrac{dfrac{sin(56^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)}=dfrac {2sin(56^circ)cos(22^circ)}{sin(34^circ)sin(56^circ)} = dfrac{2cos (22^circ)}{sin(34^circ)} $$
Since the degrees aren't the same, I could not proceed further.
trigonometry
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
$endgroup$
add a comment |
$begingroup$
$$dfrac{cot(34^circ)sin(44^circ)}{sin(22^circ)sin(56^circ)} $$
Simplify given expression.
We know that $cot alpha = dfrac{sin alpha }{cos alpha }$ and $sin(44^circ) = 2sin(22^circ)cos(22^circ)$
$$dfrac{cot(34^circ) 2sin(22^circ)cos(22^circ)}{sin(22^circ)sin(56^circ)}$$
Cancelling similar terms out
$$dfrac{cot(34^circ)2cos (22^circ)}{sin(56^circ)} = dfrac{dfrac{cos(34^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)} $$
Using the identity $cos (90^circ-alpha ) = sin (alpha)$
$$dfrac{dfrac{sin(56^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)}=dfrac {2sin(56^circ)cos(22^circ)}{sin(34^circ)sin(56^circ)} = dfrac{2cos (22^circ)}{sin(34^circ)} $$
Since the degrees aren't the same, I could not proceed further.
trigonometry
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
$endgroup$
add a comment |
$begingroup$
$$dfrac{cot(34^circ)sin(44^circ)}{sin(22^circ)sin(56^circ)} $$
Simplify given expression.
We know that $cot alpha = dfrac{sin alpha }{cos alpha }$ and $sin(44^circ) = 2sin(22^circ)cos(22^circ)$
$$dfrac{cot(34^circ) 2sin(22^circ)cos(22^circ)}{sin(22^circ)sin(56^circ)}$$
Cancelling similar terms out
$$dfrac{cot(34^circ)2cos (22^circ)}{sin(56^circ)} = dfrac{dfrac{cos(34^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)} $$
Using the identity $cos (90^circ-alpha ) = sin (alpha)$
$$dfrac{dfrac{sin(56^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)}=dfrac {2sin(56^circ)cos(22^circ)}{sin(34^circ)sin(56^circ)} = dfrac{2cos (22^circ)}{sin(34^circ)} $$
Since the degrees aren't the same, I could not proceed further.
trigonometry
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
$endgroup$
$$dfrac{cot(34^circ)sin(44^circ)}{sin(22^circ)sin(56^circ)} $$
Simplify given expression.
We know that $cot alpha = dfrac{sin alpha }{cos alpha }$ and $sin(44^circ) = 2sin(22^circ)cos(22^circ)$
$$dfrac{cot(34^circ) 2sin(22^circ)cos(22^circ)}{sin(22^circ)sin(56^circ)}$$
Cancelling similar terms out
$$dfrac{cot(34^circ)2cos (22^circ)}{sin(56^circ)} = dfrac{dfrac{cos(34^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)} $$
Using the identity $cos (90^circ-alpha ) = sin (alpha)$
$$dfrac{dfrac{sin(56^circ)}{sin(34^circ)}2cos (22^circ)}{sin(56^circ)}=dfrac {2sin(56^circ)cos(22^circ)}{sin(34^circ)sin(56^circ)} = dfrac{2cos (22^circ)}{sin(34^circ)} $$
Since the degrees aren't the same, I could not proceed further.
trigonometry
trigonometry
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
edited Dec 1 '18 at 17:01
J.G.
23.5k22237
23.5k22237
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
asked Dec 1 '18 at 16:55
HamiltonHamilton
1798
1798
add a comment |
add a comment |
1 Answer
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$begingroup$
$$frac{2sin 68^circ}{sin 34^circ}=4cos 34^circ$$
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
$endgroup$
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$$frac{2sin 68^circ}{sin 34^circ}=4cos 34^circ$$
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
$endgroup$
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
add a comment |
$begingroup$
$$frac{2sin 68^circ}{sin 34^circ}=4cos 34^circ$$
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
$endgroup$
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
add a comment |
$begingroup$
$$frac{2sin 68^circ}{sin 34^circ}=4cos 34^circ$$
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
$endgroup$
$$frac{2sin 68^circ}{sin 34^circ}=4cos 34^circ$$
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
answered Dec 1 '18 at 16:59
J.G.J.G.
23.5k22237
23.5k22237
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
add a comment |
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
$begingroup$
It was unbelieveable mistake. Thanks for reminding!
$endgroup$
– Hamilton
Dec 1 '18 at 17:02
add a comment |
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