Dividing Pythagoras through by the hypotenuse squared

What to assume they know

• Defining sine, cosine and tangent, then their three reciprocals — the six ratios, and the squaring convention that lets the identity be written without brackets
• Why enlarging the triangle leaves every ratio unchanged — that a ratio depends on the angle and not on the size, which is why dividing the lengths out loses nothing
• The extreme cases at 0° and 90°, and the ratios that stop being defined — the values at the two endpoints, which are what the identity must be checked against there
• Pythagoras theorem for a right triangle
• Dividing every term of an equation by the same non-zero quantity
• That a squared fraction is the fraction of the squares

What they should be able to do

• Distinguish an identity from an equation that holds only for particular values
• State the one relation that every right triangle satisfies among its three sides
• Carry out the division by the square of the hypotenuse and rewrite each term as a squared ratio
• State the resulting identity and the range of angles over which the chapter claims it
• Explain why the triangle's actual size has vanished from the finished statement
• Verify the identity against the tabulated values at each of the five listed angles
• Use the identity to rewrite one ratio in terms of another

Where it usually goes wrong

• "This is a new formula to memorise alongside Pythagoras." It is Pythagoras. If a student can state Pythagoras and divide, they can produce this identity from scratch.
• "It works because the special angles happen to fit." It works before any angle is chosen. The table checks are confirmations, not the reason. Show the 20-21-29 case to break the association with neat values.
• "sin²A means the sine of A squared." It means the sine, then squared. The convention puts the exponent early to avoid brackets, and the identity is unreadable if it is taken the other way.
• "You can divide by any side you like and get this." Dividing by a different side gives a different identity with a different range — that is the whole of the next topic. The hypotenuse is chosen here because it is the one side that never vanishes.
• "The identity needs a triangle, so it cannot hold at 0° or 90°." The derivation needs a triangle; the statement does not. At the endpoints it holds because the values defined in §8.3 satisfy it, which is a consistency check on those definitions.
• "Rearranging gives a plus-or-minus, so there are two answers." For an acute angle the cosine is positive, so the negative root is discarded — the same reasoning that discarded a negative length earlier in the chapter.

Questions to check understanding

• State the identity and derive it from Pythagoras in two or three lines
• Verify the identity for a stated special angle
• Given one of the sine or cosine, use the identity to find the other
• Express a named ratio in terms of the sine alone, or in terms of the cosine alone
• Simplify an expression by replacing a difference of 1 and a squared ratio
• Prove a short identity in which this one is used exactly once

Examples worth working on the board

Inputs only. Values marked verified are worked out here on the chapter's printed data.

• What an identity is (§8.4, p. 128). The section opens by settling the word: an equation earns the name when it comes out true whatever the letters stand for, and the trigonometric version says the same thing with the angle as the letter.
• Fig. 8.21 (§8.4, p. 128). A right triangle lettered ABC with the right angle at B; the drawing puts C at the lower left, B at the lower right and A at the top, with an arc marking the angle at A. Note that this is a different arrangement from Fig. 8.4. not a reflection. Set the two side by side and the drawings sit the same way round: in both, the right-angle box is at the lower right, the third vertex is directly above it, and the hypotenuse climbs from lower left to upper right. All that has changed is the lettering. B stays at the right angle; the letters A and C have traded places, so the bottom-left vertex is A in Fig. 8.4 and C here, and the top vertex is C there and A here.
• The starting relation (§8.4, p. 128, numbered (1) on the page). For that triangle the squares on AB and BC add to the square on AC. This is the only input the whole derivation has.
• The division (§8.4, p. 128). Every term is divided by the square of AC. The left side becomes the sum of two squared quotients, AB over AC and BC over AC, and the right side becomes AC over AC squared, which is 1. Verified: AB/AC is the cosine of A and BC/AC is the sine of A, straight from the §8.2 definitions and this figure's lettering, so the identity reads: the squared cosine plus the squared sine of A comes to 1. The chapter numbers it (2).
• The range (§8.4, p. 128). The chapter claims it for every A from 0° through 90° inclusive — both endpoints in. Verified from Table 8.1, p. 125: at 0° the two squares are 1 and 0; at 90° they are 0 and 1. Both sums are 1, so the inclusive claim holds. Say clearly in the explanation that the derivation only covers the angles where a triangle exists, and that the endpoints hold because the definitions adopted in §8.3 were chosen to make them hold — see Notes.
• The three interior checks (Table 8.1, p. 125). Verified: at 30° the sum is (1/2)² + (√3/2)² = 1/4 + 3/4 = 1. At 45° it is (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1. At 60° it is (√3/2)² + (1/2)² = 3/4 + 1/4 = 1.
• The ugly check (Example 3, §8.2, p. 119). The 20-21-29 triangle from earlier in the chapter, where the sine is 20/29 and the cosine is 21/29. Verified: the sum of the squares is (400 + 441)/841 = 841/841 = 1. Worth showing precisely because none of these numbers is memorable — the identity is not living off the neatness of the special angles.
• The first use (Example 9, §8.4, p. 129). Rearranging the identity gives the squared cosine as 1 minus the squared sine, hence the cosine as a square root of that; the chapter takes the positive root and marks the choice with a bracketed question, the answer being that these are acute angles and the cosine is positive there. From that the tangent and the secant follow, both expressed with sine alone underneath. Verified as a check on the shape: substituting a sine of 1/2 gives a cosine of √3/2, a tangent of 1/√3 and a secant of 2/√3, which is the 30° column of Table 8.1.
• A proof rather than a use (Example 10, §8.4, pp. 129–130). The task is to show that a particular product of three bracketed expressions collapses to 1. The route converts everything to sines and cosines, multiplies the two brackets that are a sum and a difference of 1 and the sine, and uses the identity to turn 1 minus the squared sine into the squared cosine, which then cancels the squared cosine underneath. Verified: the collapse is exact and the identity is used at precisely one step. That single step is the topic's payoff.

Figures to have open

• Fig. 8.21 redrawn as a schematic in the orientation the book prints it — right angle at B with C on the left and A at the top — the same physical layout as Fig. 8.4, with A and C swapped. Do not flip it; the risk to guard against is silently reusing Fig. 8.4's lettering, not reusing its orientation.
• A division panel: the Pythagoras equation with a horizontal rule under every term and the same squared length written beneath each, then the three quotients resolving into names. Standard schematic; this is the entire proof in one frame.
• A check strip: five small cells, one per column of Table 8.1, each showing two squares summing to 1. Standard schematic.
• The 20-21-29 triangle reused from the earlier topic's figure, so the ugly check visibly comes from a triangle the chapter itself drew.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 8 "Introduction to Trigonometry", §8.4 Trigonometric Identities, p. 128 — the definition of an identity, Fig. 8.21, the Pythagoras relation numbered (1), the division, and the identity numbered (2) with its range
• §8.4, p. 129 — Example 9, and the opening of Example 10
• §8.4, pp. 129–130 — the completion of Example 10
• Backward links inside this chapter: the six definitions in §8.2, p. 115; the squaring convention in the Note on p. 117; the numerical instance produced in Example 3, p. 119; and Table 8.1, p. 125, which supplies every check in sections 8 and 9
• The chapter summary, §8.5, p. 132, point 6, states this identity first of the three