A Trick to Memorizing Trig Particular Angle Values Desk

In calculus lessons if you end up requested to judge a trig operate at a selected angle, it’s 99.9% of the time at one of many so-called particular angles we use in our chart. Since you’re more likely to have discovered levels first I’ll embody diploma angles within the first chart, however after that, it’s gonna be radian solely.

Start by organising the desk on scratch paper as follows:

$$start{array}c theta & 0 = 0º & tfrac{pi}{6} = 30º & tfrac{pi}{4}=45º & tfrac{pi}{3}=60º & tfrac{pi}{2}=90º hlinehline sintheta &   &   &  &   &    hline costheta &   &    &    &    &    hline tantheta &    &    &    &   &    hline finish{array} $$

Then bear in mind ##sintheta## begins at zero, fill within the sample

$$start{array}c theta & 0 & tfrac{pi}{6} & tfrac{pi}{4} & tfrac{pi}{3} & tfrac{pi}{2} hlinehlinesintheta & tfrac{sqrt{0}}{2} & tfrac{sqrt{1}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{4}}{2} hlinecostheta &    &    &    &    &    hline tantheta &    &    &    &    &    hline finish{array}$$

Then do not forget that ##costheta## begins at 1 and fill within the sample

$$start{array}c theta & 0 & tfrac{pi}{6} & tfrac{pi}{4} & tfrac{pi}{3} & tfrac{pi}{2} hlinehline sintheta & tfrac{sqrt{0}}{2} & tfrac{sqrt{1}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{4}}{2} hlinecostheta & tfrac{sqrt{4}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{1}}{2} & tfrac{sqrt{0}}{2}   hlinetantheta &    &    &    &    &    hline finish{array}$$

Then bear in mind the ratio identification ##tantheta = tfrac{sintheta}{costheta}## and divide entries within the prime row (##sintheta##) by entries within the second row (##costheta##) bear in mind the ##tfrac{1}{2}## cancels to get

$$start{array}c theta & 0 & tfrac{pi}{6} & tfrac{pi}{4} & tfrac{pi}{3} & tfrac{pi}{2} hlinehlinesintheta & tfrac{sqrt{0}}{2} & tfrac{sqrt{1}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{4}}{2}   hline costheta & tfrac{sqrt{4}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{1}}{2} & tfrac{sqrt{0}}{2} hline tantheta & tfrac{sqrt{0}}{sqrt{4}} & tfrac{sqrt{1}}{sqrt{3}} & tfrac{sqrt{2}}{sqrt{2}} & tfrac{sqrt{3}}{sqrt{1}} & tfrac{sqrt{4}}{sqrt{0}} hline finish{array}$$

Now simplify,

$$start{array}c theta & 0 & tfrac{pi}{6} & tfrac{pi}{4} & tfrac{pi}{3} & tfrac{pi}{2} hlinehline sintheta & tfrac{sqrt{0}}{2}=0 & tfrac{sqrt{1}}{2}=tfrac{1}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{3}}{2} & tfrac{sqrt{4}}{2}=1  hlinecostheta & tfrac{sqrt{4}}{2}=1 & tfrac{sqrt{3}}{2} & tfrac{sqrt{2}}{2} & tfrac{sqrt{1}}{2}=tfrac{1}{2} & tfrac{sqrt{0}}{2} =0 hlinetantheta & tfrac{sqrt{0}}{sqrt{4}} =0& tfrac{sqrt{1}}{sqrt{3}}=tfrac{1}{sqrt{3}} & tfrac{sqrt{2}}{sqrt{2}} =1 & tfrac{sqrt{3}}{sqrt{1}}=sqrt{3} & tfrac{sqrt{4}}{sqrt{0}} = textual content{ und. } hline finish{array}$$

the place und. imply undefined (division by zero).

When you’re requested for the worth of one of many different three trig features, bear in mind the reciprocal identities (co-functions pair with non-co-functions)

$$csctheta =tfrac{1}{sintheta}, quad sectheta = tfrac{1}{costheta}, quad cottheta = tfrac{1}{tantheta}$$

Diagram for Pythagorean Identities

It’s handy to memorize the x, y, r definitions of the trig features, specifically (and pondering of the above diagram):

$$costheta = tfrac{x}{r}, quad sintheta = tfrac{y}{r}, quad tantheta = tfrac{y}{x}$$

$$sectheta = tfrac{r}{x}, quad csctheta = tfrac{r}{y}, quad cottheta = tfrac{x}{y}$$

We all know that the Pythagorean theorem maintain for this proper triangle, therefore ##x^2+y^2=r^2## and we’ll use this equation to generate the Pythagorean identities for the trig features:

a) divide by by ##r^2##:

##left(tfrac{x}{r}proper) ^2+left(tfrac{y}{r}proper) ^2=1 Rightarrow cos ^2theta +sin ^2theta = 1##

b) divide by by ##x^2##:

##1+left(tfrac{y}{x}proper) ^2=left(tfrac{r}{x}proper) ^2Rightarrow 1 +tan ^2theta = sec ^2theta##

c) divide by by ##y^2##:

##left(tfrac{x}{y}proper) ^2+1=left(tfrac{r}{y}proper) ^2Rightarrow cot ^2theta +1 = csc ^2theta##

It’s alway finest to derive this Desk of Trig Values from the kernel of reminiscence, solely storing in your mind:

1. ##sintheta## begins at 0 and the sample of fractions beginning at ##tfrac{sqrt{0}}{2}## and ending at ##tfrac{sqrt{4}}{2}##
2. ##costheta## begins at 1 and the identical sample of fraction as 1) solely backwards
3. ##tantheta = tfrac{sintheta}{costheta} ## and also you divide the roots from 1) by the roots from 2) (the ##tfrac{1}{2}## cancels bear in mind)