SciTech-Mathmatics-等比数列n项加总公式 + 三角函数Trigonometric Identities you must remember: 需要记住的三角函数

1. $\large S_n = a_1 \cdot \frac{1-q^n}{1-q}=\frac{a_1-a_n \cdot q}{1-q},\ when\ q \neq 1 \ and\ a_1 \neq 0 $
2. $\large S_n=n \cdot a_1,\ when\ q=1 $
3. \(\large proof\):
   \(\large \begin{array}{rrl} \\ \because & S_n &=a_1+a_2+...+a_n \\ & &=\sum_{i=2}^{n}{a_i} + a_1 \\ & q \cdot S_n&= \sum_{i=2}^{n}{a_i} + a_{n+1} \\ & &\Uparrow a_2+ a_3+...+ a_n+ a_{(n+1)} \\ & &\Uparrow a_1 \cdot q + a_2 \cdot q +...+ a_n \cdot q \\ & a_{n+1} &=a_1 \cdot q^{n} \\ \\ \therefore & S_n-q \cdot S_n &=(1-q) \cdot S_n=a_1-a_{n+1} \\ \\ \end{array}\)
   $\large S_n =a_1 \cdot \frac{1-q^n}{1-q},\ where\ q \neq 1 $

Trigonometric Identities (Revision : 1.4)

1. Trigonometric Identities you must remember

   The “big three” trigonometric identities are
   \(\large \begin{equation} \sin^{2} t + cos^{2} t = 1 \tag{1} \end{equation}\)
   \(\large \begin{equation} \sin(A + B) = \sin A \cos B + \cos A \sin B \tag{2} \end{equation}\)
   \(\large \begin{equation} \cos(A + B) = \cos A \cos B − \sin A \sin B \tag{3} \end{equation}\)

   Using these we can derive many other identities. Even if we commit the other useful identities to memory, these three will help be sure that our signs are correct, etc.

2. Two more easy identities

   From equation (1) we can generate two more identities. First, divide each term in (1) by \(cos^{2} t\) (assuming it is not zero) to obtain
   \(\large \begin{equation} \tan^{2} t + 1 = \sec^{2} t \tag{4} \end{equation}\)

   When we divide by \(sin^2 t\) (again assuming it is not zero) we get
   \(\large \begin{equation} 1 + cot^{2} t = csc^{2} t \tag{5} \end{equation}\)

3. Identities involving the difference of two angles

   From equations (2) and (3) we can get several useful identities. First, recall that
   \(\large \begin{equation} \cos(−t) = \cos t, \sin(−t) = − \sin t. \tag{6} \end{equation}\)

   From (2) we see that

   \[\large \begin{array}{rl} \sin(A − B) &= \sin(A + (−B)) \\ &= \sin A \cos(−B) + \cos A \sin(−B) \end{array}\]

   which, using the relationships in (6), reduces to
   \(\large \begin{equation} \sin(A − B) = \sin A \cos B − \cos A \sin B \tag{7} \end{equation}\)

   In a similar way, we can use equation (3) to find

   \[\large \begin{array}{rl} \cos(A − B) &= \cos(A + (−B)) \\ &= \cos A \cos(−B) − \sin A \sin(−B) \end{array}\]

   which simplifies to

   \[\large \begin{array}{rl} \cos(A − B) = \cos A \cos B + \sin A \sin B \tag{8} \end{array} \]

   Notice that by remembering the identities (2) and (3) you can easily work out the signs in these last two identities.

4. Identities involving products of sines and cosines

   If we now add equation (2) to equation (7)

   \[\large \begin{array}{rl} \\ \sin (A−B) &= \sin A \cos B − \cos A \sin B \\ +( \sin (A+B) &= \sin A \cos B + \cos A \sin B ) \\ \end{array}\]

   we find

   \[\large \begin{array}{rl} \sin(A − B) + \sin(A + B) = 2 \sin A \cos B \end{array} \]

   and dividing both sides by 2 we obtain the identity

   \[\large \begin{array}{rl} \sin A \cos B = \frac{1}{2} \sin(A − B) + \frac{1}{2} \sin(A + B) \tag{9} \end{array} \]

   In the same way we can add equations (3) and (8)

   \[\large \begin{array}{rl} \\ \cos(A−B) &= \cos A \cos B + \sin A \sin B \\ + (\cos(A+B) &= \cos A \cos B − \sin A \sin B) \\ \end{array}\]

   to get

   \[\large \begin{array}{rl} \cos (A − B) + \cos (A + B) = 2 \cos A \cos B \end{array} \]

   which can be rearranged to yield the identity

   \[\large \begin{array}{rl} \cos A \cos B = \frac{1}{2} \cos (A − B) + \frac{1}{2} \cos (A + B) \tag{10} \end{array} \]

   Suppose we wanted an identity involving \(\sin A \sin B\). We can find one by slightly modifying the last thing we did.
   Rather than adding equations (3) and (8), all we need to do is subtract equation (3) from equation (8):

   \[\large \begin{array}{rl} \cos (A−B) &= \cos A \cos B + \sin A \sin B \\ − (\cos (A+B) & =\cos A \cos B − \sin A \sin B ) \\ \end{array} \]

   This gives

   \[\large \begin{array}{rl} \cos (A − B) − \cos(A + B) = 2 \sin A \sin B \end{array} \]

   or, in the form we prefer,

   \[\large \begin{array}{rl} \sin A \sin B = \frac{1}{2} \cos (A − B) − \frac{1}{2} \cos (A + B) \tag{11} \end{array} \]

5. Double angle identities

   Now a number of easy ones. If we let \(A = B\) in equations (2) and (3) we get the two identities

   \[\large \begin{array}{rl} \sin 2A = 2\sin A \cos A \tag{12} \end{array} \]
   \[\large \begin{array}{rl} \cos 2A = cos^{2} A−sin^{2} A \tag{13} \end{array} \]

6. Identities for sine squared and cosine squared

   If we have A = B in equation (10) then we find

   \[\large \begin{array}{rl} \cos A \cos B &= \frac{1}{2} \cos (A−B) + \frac{1}{2} \cos (A+B) \\ \Rightarrow \cos^{2} A &= \frac{1}{2} \cos 0 + \frac{1}{2} \cos 2A \end{array}\]

   Simplifying this and doing the same with equation (11) we find the two identities

   \[\large \begin{array}{rl} \cos^{2} A = \frac{1}{2} (1+\cos 2A) \tag{14} \end{array} \]
   \[\large \begin{array}{rl} \sin^{2} A = \frac{1}{2} (1−\cos 2A) \tag{15} \end{array} \]

7. Identities involving tangent

   Finally, from equations (2) and (3) we can obtain an identity for $\tan (A + B):

   \[\large \begin{array}{rl} \tan (A+B)= \frac{\sin (A+B)}{\cos (A + B)} = \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B − \sin A \sin B} \end{array} \]

   Now divide numerator and denominator by cos A cos B to obtain the identity we wanted:

   \[\large \begin{array}{rl} \tan (A+B)= \frac{\tan A + \tan B}{1−\tan A \tan B} \tag{16} \end{array} \]

   We can get the identity for tan(A − B) by replacing B in (16) by −B and noting that tangent is an odd function:

   \[\large \begin{array}{rl} \tan (A-B)= \frac{\tan A - \tan B}{1+\tan A \tan B} \tag{17} \end{array} \]

8. Summary

There are many other identities that can be generated this way. In fact, the derivations above are not unique — many trigonometric identities can be obtained many different ways. The idea here is to be very familiar with a small number of identities so that you are comfortable manipulating and combining them to obtain whatever identity you need to.