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Let V be the set of all ordered triples of real numbers, and consider the following addition and scalar multiplication operations on u=(u1,u2,u3)\mathbf{u}=\left(u_{1}, u_{2}, u_{3}\right) and v=(v1,v2,v3)\mathbf{v}=\left(v_{1}, v_{2}, v_{3}\right): u+v=(u1+v1,u2+v2,u3+v3),ku=(ku1,0,0)\mathbf{u}+\mathbf{v}=\left(u_{1}+v_{1}, u_{2}+v_{2}, u_{3}+v_{3}\right), \quad k \mathbf{u}=\left(k u_{1}, 0,0\right) (a) Compute u+v\mathbf{u}+\mathbf{v} and kuk \mathbf{u} for u=(3,−2,4)\mathbf{u}=(3,-2,4), v=(1,5,−2)\mathbf{v}=(1,5,-2), and k=−1k=-1. (b) In words, explain why V is closed under addition and scalar multiplication. (c) Since the addition operation on V is the standard addition operation on R3R^{3}, certain vector space axioms hold for V because they are known to hold in R3R^{3}. (d) Show that Axioms 7, 8, and 9 hold. (e) Show that Axiom 10 fails and hence that V is not a vector space under the given operations.

Question

Starting with the addition formulas for the sine and cosine, derive the addition formula

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}

for the tangent.

Solution

Verified
Answered 1 year ago
Answered 1 year ago

We will be using the addition formulas for Sine and Cosine which are,

sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡Acos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\begin{align*} \sin(A+B)&=\sin A\cos B+\sin B\cos A \\ \cos(A+B)&=\cos A\cos B-\sin A\sin B \\ \end{align*}

Now computing the derivation of tan⁡(A+B)\tan(A+B),

tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)(Bytan⁡θ=sin⁡θcos⁡θ)=sin⁡Acos⁡B+sin⁡Bcos⁡Acos⁡Acos⁡B−sin⁡Asin⁡B(By addition formulas)\begin{align*} \tan(A+B)&=\frac{\sin(A+B)}{\cos(A+B)} &&\left(\text{By} \tan\theta=\frac{\sin\theta}{\cos\theta}\right)\\ &=\frac{\sin A\cos B+\sin B\cos A}{\cos A\cos B-\sin A\sin B} &&(\text{By addition formulas}) \\ \end{align*}

Now dividing numerator and denominator by cos⁡Acos⁡B\cos A\cos B,

tan⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡Acos⁡Acos⁡Bcos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡Acos⁡B=sin⁡Acos⁡Bcos⁡Acos⁡B+sin⁡Bcos⁡Acos⁡Acos⁡Bcos⁡Acos⁡Bcos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡Acos⁡B=sin⁡Acos⁡A+sin⁡Bcos⁡B1−sin⁡Asin⁡Bcos⁡Acos⁡B=tan⁡A+tan⁡B1−tan⁡Atan⁡B(Bytan⁡θ=sin⁡θcos⁡θ)\begin{align*} \tan(A+B)&=\frac{\dfrac{\sin A\cos B+\sin B\cos A}{\cos A\cos B}}{\dfrac{\cos A\cos B-\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\dfrac{\sin A\cos B}{\cos A\cos B}+\dfrac{\sin B\cos A}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B}-\dfrac{\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\dfrac{\sin A}{\cos A}+\dfrac{\sin B}{\cos B}}{1-\dfrac{\sin A\sin B}{\cos A\cos B}} \\ &=\frac{\tan A+\tan B}{1-\tan A\tan B}&&\left(\text{By}\tan\theta=\frac{\sin\theta}{\cos\theta}\right) \end{align*}

Hence derived.

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