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- Question 8(i, ii & iii) Exercise 8.2
- Question 8(vii, viii & ix) Exercise 8.2
- Question 8(x, xi & xii) Exercise 8.2
- Question 8(xiii, xiv & xv) Exercise 8.2
- Question 8(xvi, xvii & xviii) Exercise 8.2
- Question 8(xix, xx, xxi & xxii) Exercise 8.2
- Question 1(i, ii, iii & iv) Exercise 8.3
- Question 1(v, vi, vii & viii) Exercise 8.3
- Question 2(i, ii, iii, iv and v) Exercise 8.3
- Question 3(i, ii, iii, iv & v) Exercise 8.3
- Question 3(vi, vii, viii, ix & x) Exercise 8.3
- Question 3(xi, xii & xiii) Exercise 8.3
Fulltext results:
- Question 2(i, ii, iii, iv and v) Exercise 8.3
- ====== Question 2(i, ii, iii, iv and v) Exercise 8.3 ====== Solutions of Question 2(i, ii, iii, iv and v) of Exercise 8.3 of Unit 08: Funda... cos 20^\circ \end{align*} GOOD =====Question 2(ii)===== Rewrite the sum or difference as a product ... ><btn type="success">[[math-11-nbf:sol:unit08:ex8-3-p5|Question 3(i, ii, iii, iv & v) >]]</btn></text>
- Question 2, Review Exercise
- lign*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \th... phi)&=\frac{56}{65} \end{align*} =====Question 2(ii)===== Given that $\sin \theta=\dfrac{3}{5}, \sin ... lign*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \th... lign*} As $\theta$ is obtuse, so $\theta$ lies in II Q. This implies $\cos \theta <0$, thus $$\cos \th
- Question 7, Exercise 8.1
- a=\dfrac{4}{3}$ find \\ (i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) $\tan(\alpha+\beta)$.... rac{20}{65} \\ & = \dfrac{56}{65}. \end{align*} (ii) $\cos(\alpha + \beta)$ \begin{align*} \cos(\alph... frac{48}{65} \\ & = -\dfrac{33}{65}\end{align*} (ii) $\tan(\alpha + \beta)$ \begin{align*} \tan(\alph
- Question 8(i, ii & iii) Exercise 8.2
- ====== Question 8(i, ii & iii) Exercise 8.2 ====== Solutions of Question 8(i, ii & iii) of Exercise 8.2 of Unit 08: Fundamental of... a) \\ &=RHS \end{align*} GOOD =====Question 8(ii)===== Verify the identities: $\tan 2 x=\dfrac{1}{
- Question 1(i, ii, iii & iv) Exercise 8.3
- ====== Question 1(i, ii, iii & iv) Exercise 8.3 ====== Solutions of Question 1(i, ii, iii & iv) of Exercise 8.3 of Unit 08: Fundamenta... (26x)+\sin(6x)] \end{align*} GOOD =====Question 1(ii)===== Use the product-to-sum formula to change th
- Question 3(i, ii, iii, iv & v) Exercise 8.3
- ====== Question 3(i, ii, iii, iv & v) Exercise 8.3 ====== Solutions of Question 3(i, ii, iii, iv & v) of Exercise 8.3 of Unit 08: Fundame... )} \\ & = LHS \end{align*} GOOD =====Questio 3(ii)===== Prove the identity $\dfrac{6 \cos 8u \sin 2
- Question 5 and 6, Exercise 8.1
- \beta$ in QIII, find \\ (i) $\sin(\alpha-\beta)$ (ii) $\cos(\alpha-\beta)$ (iii) $\tan(\alpha-\beta)$.... -\frac{56}{425} =-\frac{416}{425} \end{align*} (ii) $\cos(\alpha-\beta)$ \begin{align*}\cos (\alpha-
- Question 8, Exercise 8.1
- <\beta<2 \pi$ find: \\ (i) $\csc (\alpha+\beta)$ (ii) $\sec (\alpha+\beta)$ (iii) $\cot (\alpha+\beta)... dfrac{16}{65}} \\ &= \frac{65}{16}. \end{align*} (ii) \begin{align*} \cos (\alpha + \beta) &= \cos \a
- Question 7 Exercise 8.2
- 4\alpha}{8} \end{align*} GOOD =====Question 7(ii)===== Rewrite in terms of an expression containin... "right"><btn type="success">[[math-11-nbf:sol:unit08:ex8-2-p6|Question 8(i, ii & iii) >]]</btn></text>
- Question 1, Exercise 8.1
- (2(90)-60) =-\tan 60^\circ$ GOOD ===== Question 1(ii)===== Find the value of $\cos (\alpha \pm \beta),
- Question 4, Exercise 8.1
- & = \cos 9\theta . \end{align*} ===== Question 4(ii)===== Rewrite as a single expression. $\cos 7 \th
- Question 9, Exercise 8.1
- rac{1}{5\sqrt{2}}. \end{align*} ===== Question 9(ii)===== Given $\alpha$ and $\beta$ are obtuse angle
- Question 10, Exercise 8.1
- lpha = R.H.S \end{align*} GOOD ===== Question 10(ii)===== Verify: $\cos (\pi-\alpha)=-\cos \alpha$
- Question 11, Exercise 8.1
- = 1 = R.H.S \end{align*} GOOD ===== Question 11(ii)===== Show that: $\dfrac{\sin \left(90^{\circ}+\a
- Question 12, Exercise 8.1
- nd{align*} as required. GOOD ===== Question 12(ii)===== If $\alpha+\beta+\gamma=180^{\circ}$, prove