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Question 6, Exercise 10.2
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c{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identit... heta }{2}=\dfrac{{{135}^{\circ }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half angle ident... heta }{2}=\dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identi... theta }{2}=\dfrac{\dfrac{\pi }{4}}{2}$, so we can find $\cos \dfrac{\pi }{8}$by using half angle identit
Question 3, Exercise 10.2
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l ray of $\theta$ is in the second quadrant, then find $\sin2\theta$. ====Solution==== Given: $\sin \the... pk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the... l ray of $\theta$ is in the second quadrant, then find $\cos \dfrac{\theta }{2}$. ====Solution==== Given... pk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have th
Question, Exercise 10.1
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pha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \r... pha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\cos \left( \alpha +\beta \r... pha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\tan \left( \alpha +\beta \
Question 5, Exercise 10.1
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$\alpha$ nor $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solut... $\alpha$ nor $\beta$ in the first Quadrant, then find: $\cos \left( \alpha +\beta \right)$. ====Solut... $\alpha$ nor $\beta$ in the first Quadrant, then find: $\tan \left( \alpha +\beta \right)$. ====Solut
Question 2, Exercise 10.2
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ray of $\theta $ is in the second quadrant, then find $\sin 2\theta $. ====Solution==== Given: $\sin \... ray of $\theta $ is in the second quadrant, then find $\cos 2\theta $. ====Solution==== Given: $\sin \... ray of $\theta $ is in the second quadrant, then find $\tan 2\theta $. ====Solution==== Given: $\sin \
Question 1, Exercise 10.2
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re four parts in Question 1. =====Question1===== Find the value of $\sin 2\theta ,\,\,\cos 2\theta$ and... k-ex10-2-q1.png?nolink |reference triangle}} we find $\sin \theta =\dfrac{1}{\sqrt{26}}$ and $\cos \
Question 4 and 5, Exercise 10.2
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l ray of $\theta $ is in the third quadrant, then find $\sin \dfrac{\theta }{2}$. ====Solution==== Given