I was reading a trig book in which the author had deduced the following formulas from a drawing. I am mentioning those as follows...

enter image description here

By looking at the above drawing the author deduced these formulas for $(\theta+90^\circ)$...

1. $\sin (\theta + 90^\circ)=\cos\theta$

2. $\cos (\theta + 90^\circ)=-\sin\theta$

3. $\tan (\theta + 90^\circ)=-\cot\theta$

I understood how did he conclude above formulas from the drawing.

As you can see the drawing depicts author's approach to derive the $(\theta+90^\circ)$ formulas by taking a similar triangle in the second quadrant, in the same way, to derive $(\theta-90^\circ)$ formulas, he advised to take a similar triangle (as in the drawing) in quadrant 4 as well. But, to the contrary he mentioned a trick to get those formulas which I do not understand.

The author told to replace $\theta$ in formulas 1,2 and 3 by $(\theta-90^\circ)$. So what I got by doing that is...

1. $\sin ((\theta-90^\circ) + 90^\circ)=\cos(\theta-90^\circ)$

2. $\cos ((\theta-90^\circ) + 90^\circ)=-\sin(\theta-90^\circ)$

3. $\tan ((\theta-90^\circ) + 90^\circ)=-\cot(\theta-90^\circ)$

By evaluating them further I get...

1. $\sin (\theta)=\cos(\theta-90^\circ)$ (I got this one correct)

2. $\cos (\theta)=-\sin(\theta-90^\circ)$ (wrong)

3. $\tan (\theta)=-\cot(\theta-90^\circ)$ (wrong)

But the author says that the outcome of the formulas should be...

1.$\sin (\theta)=\cos(\theta-90^\circ)$ (correct.)

2. $-\cos (\theta)=\sin(\theta-90^\circ)$

3. $\tan (\theta-90^\circ) =-\cot(\theta)$

I don't know where am I going wrong. The author told to replace $\theta$ by $(\theta-90^\circ)$ then why am I getting my answers incorrect? Where is the flaw? Please help me to track it down.

  • 2
    $\begingroup$ Just because you don't get the exact character-for-character representation of the equation, doesn't mean you did it wrong. For example, the (2) that you have is equivalent to the (2) the author has. Can you see why? $\endgroup$ – Henry Swanson Nov 15 '17 at 7:15
  • $\begingroup$ The $-$ sign is on opposite sides in both those equations. How is it correect? I can't see why. $\endgroup$ – Saksham Sharma Nov 15 '17 at 7:17
  • 1
    $\begingroup$ @SakshamSharma Assume $u=-v$. Multiply both sides by $(-1)$ to obtain $-u=v$. $\endgroup$ – Hagen von Eitzen Nov 15 '17 at 7:20
  • $\begingroup$ @HagenvonEitzen I understood. Your convention holds true for $2$. What about the last one? $3$? $\endgroup$ – Saksham Sharma Nov 15 '17 at 7:22
  • 1
    $\begingroup$ Well, you have $\tan(\mathrm{thing}) = -\cot(\mathrm{other thing})$. What do you know about $\tan(\mathrm{other thing})$ and $\cot(\mathrm{thing})$? $\endgroup$ – Henry Swanson Nov 15 '17 at 8:15

See, for equation 2 it is simple. Assume you have..

$u=-v$ then you can get $-u=v$ by multiplying the equation on both sides by $-1$.

For 3, use the reciprocal identity...$$\tan(x)=\frac{1}{\cot(x)}$$

and substitute it in the respective variables to arrive at the desired answer.


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