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- Question 8 & 9, Review Exercise 10
- ====== Question 8 & 9, Review Exercise 10 ====== Solutions of Question 8 & 9 of Review Exercise 10 of Unit 1... }+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2\sin \alpha \sin \beta =\cos \... c\left( 2\pi -\theta \right)}=\cos \theta $. ====Solution==== \begin{align}L.H.S.&=\dfrac{{{\sin }^{2}}\lef... \left( {{90}^{\circ }}-\theta \right)}=-1$. ====Solution==== \begin{align}L.H.S.&=\dfrac{\cos \left( {{90}
- Question 6 & 7, Review Exercise 10
- ====== Question 6 & 7, Review Exercise 10 ====== Solutions of Question 6 & 7 of Review Exercise 10 of Unit 1... 1-8{{\sin }^{2}}\theta {{\cos }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\cos 4\theta \\ &=\cos 2\... \sin 6x\sin x+\cos 4x\cos 3x=\cos 3x\cos 2x$. ====Solution==== \begin{align}L.H.S.&=\sin 6x\sin x+\cos 4x\co
- Question 4 & 5, Review Exercise 10
- ====== Question 4 & 5, Review Exercise 10 ====== Solutions of Question 4 & 5 of Review Exercise 10 of Unit 1... rac{\sin \theta \tan \dfrac{\theta }{2}}{2}$. ====Solution==== \begin{align}R.H.S.&=\dfrac{\sin \theta \tan... heta \tan \dfrac{\theta }{2}=\sec \theta -1$. ====Solution==== \begin{align}L.H.S.&=\tan \theta \tan \dfrac{
- Question 1, Review Exercise 10
- ====== Question 1, Review Exercise 10 ====== Solutions of Question 1 of Review Exercise 10 of Unit 10: Trig
- Question 2 and 3, Review Exercise 10
- ===== Question 2 and 3, Review Exercise 10 ====== Solutions of Question 2 and 3 of Review Exercise 10 of Uni... a +\cos 3\theta }=\tan 2\theta \tan \theta $. ====Solution==== \begin{align}L.H.S.&=\dfrac{2\sin \theta \si... a}{\sin 4a+\sin 2a}=\dfrac{\cos 7a}{\cos a}$. ====Solution==== \begin{align}L.H.S.&=\dfrac{\sin 10a-\sin 4a
- Question 8, Exercise 10.1
- ====== Question 8, Exercise 10.1 ====== Solutions of Question 8 of Exercise 10.1 of Unit 10: Trigonometric ... a +\sin \theta }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... \right)=\dfrac{1-tan\theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi ... {1-{{\tan }^{2}}\alpha {{\tan }^{2}}\beta }$ ====Solution==== \begin{align}\tan (\alpha +\beta )&=\dfrac{\
- Question 2, Exercise 10.1
- ====== Question 2, Exercise 10.1 ====== Solutions of Question 2 of Exercise 10.1 of Unit 10: Trigonometric ... === Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }... ii)=== Evaluate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }... i)=== Evaluate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^
- Question 5, Exercise 10.3
- ====== Question 5, Exercise 10.3 ====== Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric I... {\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos... \pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos... {\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\sin \alpha \sin \beta =\cos
- Question 5, Exercise 10.3
- ====== Question 5, Exercise 10.3 ====== Solutions of Question 5 of Exercise 10.3 of Unit 10: Trigonometric I... \circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos... pi }{3}\sin \dfrac{4\pi }{9}=\dfrac{3}{16}.$$ ====Solution==== We have an identities: $$2\sin \alpha \sin \b... {\circ }}\sin {{70}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We have an identities: $$2\sin \alpha \sin \b
- Question 3, Exercise 10.3
- ====== Question 3, Exercise 10.3 ====== Solutions of Question 3 of Exercise 10.3 of Unit 10: Trigonometric I... }^{\circ }}-\sin {{15}^{\circ }}}=\sqrt{3}.$$ ====Solution==== We have identities: $$\cos \alpha +\cos \beta... circ }}+\cos {{120}^{\circ }}}=3+2\sqrt{2}.$$ ====Solution==== \begin{align}L.H.S.&=\dfrac{\sin {{135}^{\ci
- Question 2, Exercise 10.3
- ====== Question 2, Exercise 10.3 ====== Solutions of Question 2 of Exercise 10.3 of Unit 10: Trigonometric I... $\sin {{37}^{\circ }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \be... $\cos {{36}^{\circ }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \be... $$\sin \dfrac{P+Q}{2}-\sin \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$\sin \alpha -\sin \be
- Question 1, Exercise 10.3
- ====== Question 1, Exercise 10.3 ====== Solutions of Question 1 of Exercise 10.3 of Unit 10: Trigonometric ... roduct as sum or difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \b... $\sin {{55}^{\circ }}\cos {{123}^{\circ }}$. ====Solution==== We have an identity: $$2\sin \alpha \cos \b... : $$\sin \dfrac{A+B}{2}\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$2\sin \alpha \cos \
- Question 8 and 9, Exercise 10.2
- ====== Question 8 and 9, Exercise 10.2 ====== Solutions of Question 8 and 9 of Exercise 10.2 of Unit 10: Tri... first power of one or more cosine functions. ====Solution==== \begin{align}{{\cos}^{4}}\theta &={{\left( {{... os }^{3}}\theta -4\sin \theta \cos \theta $ . ====Solution==== \begin{align}L.H.S.&=\sin 4\theta \\ &=\sin 2... c{1-{{\tan }^{2}}2\theta }{2\tan 2\theta }$ . ====Solution==== \begin{align}L.H.S.&=\cot 4\theta =\dfrac{\co
- Question 7, Exercise 10.2
- ====== Question 7, Exercise 10.2 ====== Solutions of Question 7 of Exercise 10.2 of Unit 10: Trigonometric I... \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1
- Question 4 and 5, Exercise 10.2
- ====== Question 4 and 5, Exercise 10.2 ====== Solutions of Question 4 and 5 of Exercise 10.2 of Unit 10: Tri... uadrant, then find $\sin \dfrac{\theta }{2}$. ====Solution==== Given: $\cos \theta =-\dfrac{3}{7}$ and termi... to evaluate exactly $\sin \dfrac{2\pi }{3}$. ====Solution==== Given: $\sin \dfrac{2\pi }{3}$.\\ By using do... to evaluate exactly $\cos \dfrac{2\pi }{3}$. ====Solution==== Given: $\cos \dfrac{2\pi }{3}$ \\ By using do